Programming Analysis In C (2E)

PROGRAMMING

ANALYSIS IN

C

PROGRAMMING

ANALYSIS IN

C

Dedicated To,

My My My My Late Grandfather Late Grandfather Late Grandfather Late Grandfather Mitra Man PalikheMitra Man PalikheMitra Man PalikheMitra Man Palikhe

and also to School Of Engineering (SOE)-POKHARA

UNIVERSITY & PALIKHE FAMILY

2069 Asoj 16/2012 October 2 ‘TUE’

Devendra Man Palikhe

([email protected])

You can get soft copy of this book @ www.scribd.com/dpalikhe

PREFACE

This book entitled “PROGRAMMING ANALYSIS IN C” on C-Programming is the first

of its kind being published and written by student of SCHOOL OF ENGINEERING

(SOE), POKHARA UNIVERSITY. It mainly presents the programs and their solutions

regarding C-Programming during first semester in Engineering.

This book is meant to help the reader learn, understand, familiarize, and analyze

how to program in C. This book is for the purpose of building up the creative

ideas regarding C-Programming in mind of learners instead of the definitions

and it includes all the difficulties that the learners encounter during their voyage

in C-Programming.

This book is mainly meant for the difficult programs that learners come across

their journey in C-Programming and it includes all those questions which have

appeared in previous examinations of POKHARA UNIVERSITY that are a hard nut

to crack for the students but not merely all the questions. Most of the tough

solution of this book is based on analyzing problem through possible illustrative

ways, analyzing the problem for coding it into programs rather than on plain

statements of rules.

For the majority part, the programs are entire, real programs, rather than solitary

fragments. So this book is not an opening programming manual; it assumes some

experience and knowledge with programming concepts like variables,

assignments, statements, loops, logics, and functions.

The small programs, at last in projects that are just to enhance the creativity of

the learners, are self understood as they are written in very simple statements

and in sequential manner with outputs. All programs of this book are complied in

‘Borland C++ 5.02’.

I wish to acknowledge my greatest indebtness to my teacher Yagyanath Rimal

and my brothers Nitesh Palikhe and Nischal Palikhe for their constructive

suggestions, encouragement and constant support during publication of this

book. I would also like to thank all my colleagues. All the thoughtful criticisms and

critical suggestions of my friends, colleagues, readers and well-wishers for

improvement of this book are heartily welcome and will be highly appreciated.

ACKNOWLEDGEMENT

My sincere gratitude goes to my colleagues of school of engineering whose

constant support and encouragement has always been my source of

inspiration.

Devendra Man Palikhe

2�� Semester, BCE

School Of Engineering

POKHARA UNIVERSITY

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CONTENTS

SOME BEGINNING PROGRAMS

Decision making

Loop A program to convert degree Celsius to Fahrenheit

A program to find positive or negative

A program to find numbers between 0 to 100 divisible by 3 and not by 5

Factorial`

Leap year

H.C.F. by division method

Reference-interchange two values

Flowchart of

Greatest among four numbers

Greatest among three numbers

Some types of numbers

Armstrong numbers

Even and odd numbers

Prime numbers

FIBONACCI SERIES

1, 1, 2, 3, 5, 8, 13, 21 …………….……….

0, 1, 1, 2, 3, 5, 8, 13, 21 ………………

DIGITS OF A NUMBER

Digits taking from a number

PALINDROME

Palindrome for number

An analysis of returning back number from its digits

Palindrome for string

Using reversing string function A char[] is declared where each time a new last character from string is taken one by one as array and then check is done to original string Palindrome for string containing even or odd quantity of characters Palindrome for string using for-loop

ELECRTICITY AUTHORITY

ARRAY

2d

PU2008FALL 2b) finding an element of 2D-array

Finding greatest and smallest number among ‘n’ numbers (using array)

PATTERNS OF CHARACTERS AND NUMBERS

‘Scanf’ function for numbers

Patterns of characters and numbers using ‘printf’ function

Basic principle of printing patterns

Some more patterns

Floyd’s triangle

Patterns of ‘*’

Some more difficult patterns

Pyramid of triangle

Reverse pyramid of triangle

Pascal’s triangle

PU2002 FALL 5 WAP that takes angle in degrees and give out value of sine for that angle without using ‘sin ()’ function of math library

Conversion

Decimal to binary

Binary to decimal

Exclusive use of ‘switch-case’ and ‘goto’ statements

2005PU A program that takes a sum of money and converts it into smaller units, where user can choose his starting unit. The conversion is in such a way that the first starting units takes maximum possible conversion then the left sum is supplied to conjugate term following it, this process continues till the end

SORTING

Selection sorting

Bubble sorting

Third sorting process

Analysis of the solution of third sorting

A special analysis in sorting-third sorting

PU2011FALL 5b) Sort ‘n’ numbers by function (special)

Description

A SPECIAL ANALYSIS

Now we are adding a new feature to above program

Examples of sorting of numbers and strings in selection and bubble sorting

Median and standard deviation

MATRIX ARRAY

Case-1: Finding smallest and greatest of a matrix

Case-2: Sum of diagonal elements of square 3*3 matrix

1. sum of diagonal from right top to left bottom 2. sum of diagonal from left top to right bottom

Case-3: Finding a matrix element in a matrix

Case-4: sum of all elements in a matrix

Case-5: transpose of a matrix

Matrix Multiplication

Matrices multiplication of (3*3) and (3*3)

More generalization of matrices multiplication

TASK A simple program that takes students marks in various subjects, name and roll numbers to print out ranks, roll numbers and marks (both % and marks) in Descending order.

Solution of the task

RECURSIVE FUNCTION

Factorial by function

Factorial by recursive

Fibonacci series by recursive

1 1 2 3 5 8 13 21 ……………….


0 1 1 2 3 5 8 13 21 ……………….

To calculate power by recursive

Calculate sum of digits by recursive

Armstrong using function

Armstrong using recursive

WITHOUT FUNCTION

To check case without using any function

Calculate power without function

To reverse a sentence without any function

Concatenate two strings without function

A DETAIL OF STRINGS

Finding a sequence of words

Finding a sequence of words any

Finding a sequence of words (working)

Finding a word in a sentence

Finding a word in a sentence (working)

Finding a word in a sentence (input from user)

OPERATORS

Table

Post/pre increment/decrement operators

For-block’s execution flow

ASSIGNMENT OPERATORS

OUTPUTS

Hints to PU 2008 FALL 3a) 3b

PU 2006 SPRING 2 a) 2 b) 2 c)

PU 2009 SPRING 2 b) important 3 a) very important

PU 2011 FALL 3 a) 3 b)

PU 2009 FALL 3 a) 3 b)

PU 2007 FALL 2 a) 2 b) PU 2012 FALL 2b i) very important

PU 2011 SPRING Ternary operator’s nested use

PU2006FALL 3) Print out combinations of 1, 2, 3 using for-loop

PUFALL2003 2 b) Read temperature of a month of 10 cities and print out average of ten cities using array and for-loop

Second way of solving

POINTERS

Pointer declaration in arrays

Program demonstrating pointer variable’s resembling with special array i.e. character array (string)

Using ‘void’ type pointer as required

Demonstrating the 2d pointers in 2d array

Call by address/ pass by pointer

Pointers in structure

Passing whole structure to function

PU 2006 SPRING 5a) Meaning of some pointer expressions

2003 fall 4 a) Write a program with user defined function using pointers to convert all the uppercase letters to lower-case and vice versa in a string given by the user

SOME CONCEPTUAL TERMS

Returning multiple value by functions

Program documentation

Important casting mechanism

Variables

Call by value vs. call by reference

A simple program demonstrating structure: It takes data and print info of whose address is “pokhara”

FILE HANDLING

File pointer vs. std

PU2009 SPRING 6 b) WAP to read name, author and price of 500 books and store the info to “library.dat” and print info of books of price > 300

PU2009 FALL 6 b) Create a structure for the following data: Id, Name, Address, Salary, Date of join and write program to input 100 employee, save in emp.dat file then print info of employee who have the address 'pokhara'

Due fee Prints student info whose due fee is less than 100

PU2002 Consumed charged data merged with file handling: WAP that stores following data of customer of electricity authority in file ‘elect.dat’: id, name, old reading, new reading and amount to be paid

PROJECTS

Books

Phone book

Mark sheet by harder method

Mark sheet by easier method

Updated mark sheet

Stores

Updated stores




PROGRAMMING ANALYSIS IN C

By Devendra Man Palikhe

1

SOME BEGINNING PROGRAMS

DECISION MAKING

While going to the solution of a problem, we encounter many problems to which small decisions or a

group of small decisions or a loop of small decision can give a meaningful output. As much as

possible, decisions must be easier, simpler to read, understand and analyze being in great purpose

to the problem’s solution.

LOOP

One of the important processes in C programming is the repetition of statements (same for all time

or change gradually with time or condition or variables or inputs supplied by users, etc.). The array is

based on this looping (especially dynamic array).

A PROGRAM TO CONVERT DEGREE CELCIUS TO FAHRENHEIT

//A PROGRAM TO CONVERT DEGREE CELCIUS TO FAHRENHEIT

#include<stdio.h>

#include<conio.h>

int main() {

int celcius, fahrenheit;

clrscr();

printf("Enter temperature in celcius\t");

scanf("%d",&celcius);

fahrenheit=(celcius*1.8)+32;

printf("\n The equivalent temperature of %d in Fahrenheit is %d",celcius,fahrenheit);

getch();

return 1;}

A PROGRAM TO FIND POSITIVE OR NEGATIVE

We can conclude a number is positive or negative by comparing it with zero and the sum here is the

previous value stored in sum plus new value need to be stored in ‘sum’ i.e. ‘sum=sum + num’. Here

sum is calculated and average is calculated by dividing sum by number of items. To calculate the

number of items we declared a counter whose initial value is zero and it gets incremented by one

each time the second if gets executed and hence average is calculated.

#include<stdio.h>

#include<conio.h>

void main() {

int n,num[100],positive=0,negative=0,i;

printf("Enter how many numbers do you want to test?");

scanf("%d",&n);

for(i=0;i<n;i++) {

printf("Enter a number\n");

scanf("%d",&num[i]);

if(num[i]>0) {

positive=positive+1;

}



2

else if(num[i]<0) {

negative=negative+1;

}

else {

// for the case of zero number we do nothing

}

}

printf("\nThe number of positive numbers is %d",positive);

printf("\nThe number of negative numbers is %d",negative);

getch();}

A PROGRAM TO FIND NUMBERS BETWEEN 0 TO 100 DIVISIBLE BY 3 AND NOT BY 5

#include<stdio.h>

#include<conio.h>

void main() {

int i;

printf("The numbers between 0 to 100 that are divisible by 3 exactly and not divisible by 5 are:");

for(i=1;i<=100;i++) {

if(i%3==0 && i%5!=0)

printf("\n%d",i);

}

getch();}

Here numbers are taken from 0 to 100 by the help of for-loop (pre-test loop) and others are

arithmetic and logical operations where ‘%’ modulo sign gives remainder.

We can extend this program by asking n1 and n2 and fixing these limits for the values of i, so that

only numbers from n1 to n2 are checked where n1<n2.

FACTORIAL

n!=1*2*3*4*5……….(n-2)*(n-1)*n

#include<stdio.h>

#include<conio.h>

void main() {

int i,factorial,n;

factorial=1;

printf("Enter number?");

scanf("%d",&n);

for(i=1;i<=n;i++)

factorial=factorial*i;

printf("\nThe required factorial is%d",factorial);

getch();}

Here the factorial is found by multiplying from 1 to n. For this initialization is made factorial=1 (it

simply represents factorial of 1) and then the loop of i (from 1 to n) is taken which multiply the



3

previous factorial by a new number i (‘i’ is incremented each time by one to final to n) to get factorial

of new number ‘i’ i.e. a new value of i. So factorial at initial is initialized 1 so that multiplication runs

smoothly, it should be clear that initializing zero creates error.

LEAP YEAR

Firstly a leap year is exactly divisible by 4. A year greater than 100 must be divisible by 4 is leap. A

year less than 100 but divisible by 400 and 4 is leap year. It is because for number smaller than 100;

if it is divisible by 4 then it is leap year but for the number greater than 100 (i.e. takes place between

100- 1000), we calculate by division with 40 also so that 400 (is 4’s multiple in 100’s range i.e. 100-

1000) if is exactly gone, is leap for 100’s (i.e. 100-1000) range else not.

#include<stdio.h>

#include<conio.h>

void main() {

int year;

int factorial=1;

printf("Enter a year?");

scanf("%d",&year);

if(year%4==0){

if(year%100!=0){

printf("\nThe given year is a leap year");

}

else if(year%400==0){

printf("\nThe given year is a leap year");

}

else {

printf("\nThe given year is not a leap year");

}

}

else {

printf("\nThe given year is not a leap year");

}

getch();}

H.C.F BY (DIVISION METHOD)

Here in between two numbers (or can be extended to more numbers), a number is divided by

another number; the remainder divides the divisor till the remainder is zero (for more than two

numbers, each new number is added in dividend and above process continues till remainder zero).

For the last instant, the divisor is the H.C.F. (Highest Common Factor).

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4

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

void main() {

int n1,n2,divisor,divident,remainder;

printf("Enter First Number");

scanf("%d",&n1);

printf("Enter Second Number");

scanf("%d",&n2);

if(n1>n2){

divident=n1;

divisor=n2;

}

else if(n2>n1){

divident=n2;

divisor=n1;

}

else if(n2=n1){

printf("The H.C.F. of two numbers is %d",n1);

exit(0);

}

do {

remainder=divident%divisor;

divident=divisor;

divisor=remainder;

}

while(remainder>0);

printf("The H.C.F. of two numbers is %d",divident);

getch();}



5

REFERENCE- INTERCHANGE TWO VALUES

Two variables have two values. In some other languages like Q-Basic, a direct function ‘swap’

interchanges their values regarding to identifiers. However there is no such in C. For e.g. if we

directly declare an identifier to value of other identifier i.e. value_a =value_b

Then both will have value b against our intension to also to get value_b=a, but here value_a gets b

and its original value ‘a’ disappears which can’t be restored. So a temporary variable must store the

value. Take two stone in two hands. How can they be interchanged, one is directly but in C there is

not so function so the other indirect method is used. Here the stone is transferred from one hand to

table and then the stone of other hand is transferred to first hand and then stone from table to

second hand. This is similar to declaring temporary variable and storing a value whose identifier is

made to take other value and then assigning the temporary value to the identifier whose value was

assigned by other. Thus declaration completes.

temp =value_b;

value_b=value_a;

value_a=temp;

#include<stdio.h>

#include<conio.h>

void main() {

int diff,a,b;

printf("Enter a Number");

scanf("%d",&a);

printf("\nEnter another Number");

scanf("%d",&b);

diff=a-b;

a=a-diff;

b=b+diff;

printf("\nThe interchanged numbers are like of this order %d\t%d",a,b);

getch();}

The next method is to take difference between two numbers taking one as a reference. The

difference is according to the reference number (the number before minus) so the number taken as

reference is subtracted to difference to get another number (because reference is the comparison of

another number with respect to a number) and another is added to difference to get first number as

its comparison s already with reference number. Making a number reference is to compare

according to its magnitude or direction but here magnitude is sufficient.

#include<stdio.h>

#include<conio.h>

void main() {

int diff,a,b;


scanf("%d",&a);

printf("\nEnter another Number");

scanf("%d",&b);

diff=a-b;

a=a-diff;



6

b=b+diff;

printf("\nThe interchanged numbers are like of this order %d\t%d",a,b);

getch();}

FLOW CHART

Greatest among four numbers

Greatest among three numbers



7

SOME TYPES OF NUMBERS

ARMSTRONG NUMBERS

It is number equal to sum of cubes of all the digits present in number. For e.g. 153, 370, 371.

To check Armstrong, take each digit and save each time to sum its cube (i.e. Sum=sum + digit’s

power 3) and at last check the sum to original number. Please refer ‘digit taking from a number’

given in coming part; for the digit taking process.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main() {

int num,digit,sum=0,temp_num;

clrscr();

printf("Enter a number");

scanf("%d",&num);

temp_num=num; /*here temp_num is necessary to hold transition values of num*/

while(temp_num>0){

digit= temp_num%10;

temp_num=temp_num/10;

sum=sum+pow(digit,3);

}

if(sum==num) printf("IT IS ARMSTRONG"); /*to preserve originality necessary for comparison

'num' holds original value */

else printf("IT IS NOT ARMSTRONG");

getch();

}

Please look ‘recursive function’ for further advancement.

EVEN AND ODD NUMBERS

Find remainder by division by two using “%” (modulo division) and then and check it is 0 or not, if 0

then even else odd.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main() {

int num;

clrscr();


scanf("%d",&num);

if(num%2==0) printf("IT IS EVEN");

else printf("IT IS ODD");

getch();

}



8

We generally use ‘for-loop’ to add limiting values for numbers to be checked. Let’s us look a small

program that prints even numbers from 107 to 221 and prints sum of odd numbers of the given

range.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main() {

int odd_sum=0,i;

printf("The even sumbers are\n");

for(i=107;i<=221;i++){

if(i%2==0) printf("\t%d",i);

else odd_sum=odd_sum+i; /*those which are not even are odd*/

}// for-loop terminal

printf("\nThe sum of odd is:\t%d",odd_sum);

getch();

}

PRIME NUMBERS

Prime number is such which is only exactly divisible by 1 or itself but not by others i.e. modulo

division (%) by only 1 and itself gives zero remainder but not by any else.

#include<stdio.h>

#include<conio.h>

#include<stdlib.h> //for ‘exit()’function

void main() {

int number,i;

clrscr();


scanf("%d",&number);

for(i=2;i<number;i++) {

if(number%i==0) {printf("NOT PRIME NUMBER"); if (getch()) exit(0); }

}

printf("IT IS PRIME NUMBER");

getch();

}

Here we are checking the division of the number from 2 to a value less than number, if it is divisible

for at least for one case then “it is not even” or “it is composite” but if the whole ‘for-loop’

terminates without any possible remainder zero then at last “it is prime” is printed. It must be clear

that ‘exit()’ function jumps all execution flow to end of program escaping all statements directly.

#include<stdio.h>

#include<conio.h>

void main() {

int number,i,divisible=0; /*here for others like number and i initialization is not compulsory as i is

intialized in for-loop and for number it is initialized by 'scanf-function'*/

clrscr();



for(i=1;i<=number;i++) {



9

if(number%i==0) divisible++;

}

if(divisible==2) printf("IT IS PRIME NUMBER");

else printf("IT IS COMPOSITE OR NOT PRIME NUMBER");

getch();

}

Here ‘i’ loop starts from zero to number and for each exact division, variable ‘divisible’ is

incremented by one, so that after the for-loop terminates if ‘divisible’ is equal to two only; shows

that the number was divisible by one and itself only but not by any numbers from zero to number,

so it is prime.

NOTE: Every number is exactly divisible by one and by itself, prime numbers are strictly limited to it

but composite numbers are exactly divisible by at least one new number in addition to it.

#include<stdio.h>

#include<conio.h>

void main() {

int number,divident=2;

clrscr();



while(divident<number){

if(number%divident==0) { printf("IT IS COMPOSITE OR NOT PRIME NUMBER"); if(getch()) exit(0); }

/*here getch() is used to hold statement printed on the screen beefore 'exit'-function is executed */

divident++;

}


getch(); // to hold the value printed on screen till any character is given

}

Here we are using ‘while’-loop instead of ‘for’-loop. It is similar to the case of ‘for’-loop. These ‘for-

loop’, ‘while-loop’ can also be replaced by ‘goto-statement’ like this

#include<stdio.h>

#include<conio.h>

void main() {

int number,divident=2;

clrscr();



again:

if(divident<number) { /* here we are limiting the range of divident, it is our first aim to limit the

divident than the checking of divisible so this 'if' is used at first*/

if(number%divident==0) { printf("IT IS COMPOSITE OR NOT PRIME NUMBER"); if(getch()) exit(0); }

divident++; goto again; /*we can place these statements in 'else'-statement too*/

}


getch();

}



10

FIBONACCI SERIES

IF THE FIBONACCI SERIES IS

1, 1, 2, 3, 5, 8, 13, 21 …………….……….

We see pattern from third term that it is sum of two numbers before it; 2=1+1, 3=2+1, 5=3+2……….

For this, our result is sum of two numbers whose one number is previous sum and other is one step

more previous sum (or the number proceeding to what we are concerning to for sum or the number

we are concerning with).

a b c printed value

(c) statement value statement value statement value

a=0 0 b=1 1 a+b 1 1

a=b 1 b=c 1 a+b 2 2

a=b 1 b=c 2 a+b 3 3

a=b 2 b=c 3 a+b 5 5

a=b 3 b=c 5 a+b 8 8

a=b 5 b=c 8 a+b 13 13

a=b 8 b=c 13 a+b 21 21

a=b 13 b=c 21 a+b 34 34

And this process goes on. In the program ‘1’ at initial is printed to maintain pattern and this

happened due to inefficient initialization of values ‘a’ and ‘b’. The approaching next method cracks

it.

#include<stdio.h>

#include<conio.h>

void main() {

int a=0;

int b=1;

int c,i,n;

printf("Enter a Number\t");

scanf("%d",&n);

printf("%d\t",b);

for(i=1;i<n;i++) {

c=a+b;

printf("%d\t",c);

a=b;

b=c;

}

getch();}

Here for our initialization purpose we declare a=0, b=1 so that our sum c=0+1 is 1 initially and

second time after interchanging a, b, c we get c=2 so we print b (=1) initially we continue the pattern

however the loop continues for n-1 times as the first number is made to print automatically on any

cost or input.

IF THE FIBONACCI SERIES IS



11

0, 1, 1, 2, 3, 5, 8, 13, 21 ………………

Then we conclude that 0 must be sum of two terms in such a way that the second of two terms=zero

=1. So these numbers must be -1 and 1. Other interchanging and printing process are same to above.

a b c printed value

(c) statement value statement value statement value

a=-1 -1 b=1 1 a+b 0 0

a=b 1 b=c 0 a+b 1 1

a=b 0 b=c 1 a+b 1 1

a=b 1 b=c 1 a+b 2 2

a=b 1 b=c 2 a+b 3 3

a=b 2 b=c 3 a+b 5 5

a=b 3 b=c 5 a+b 8 8

a=b 5 b=c 8 a+b 13 13

And this process goes on.

#include<stdio.h>

#include<conio.h>

void main() {

int a=-1,b=1,c,n,i;

printf("Enter a number\t");

scanf("%d",&n);

for(i=1;i<=n;i++) {

c=a+b;

printf("%d\t",c);

a=b;

b=c;

}

getch();}



12

DIGITS OF A NUMBER

DIGITS TAKING FROM A NUMBER

Digit taking from a number is very useful and important task because while dealing to numbers in C,

we need the digits’ calculations. Digits of any number in its corresponding number system are

nothing but the remainder by division by the base of the system to the particular number. So we

divide by 10 and take the remainder for the purpose. However after a digit is taken, we need to

remove a digit (last) from the number so that we can take next digit by similar above process. For

removing last digit (i.e. at one’s place) we make integer division by 10(this takes quotient before

decimal point). This process continues till num>0 and digits are taken each time to which we can

process.

#include<stdio.h>

#include<conio.h>

void main() {

int digit[15],num,i=0,j;


scanf("%d",&num);

while(num>0) {

digit[i]=num%10;

num=num/10;

i=i+1;

}

for(j=i-1;j>-1;j--){

printf("\nThe digit in %d place is %d",j+1,digit[j]);

}

getch();}

The next way is not to remove digit(last) from numbers instead calculating digits by dividing the num

each time by one’s…..ten’s….hundred’s ………(ten’s multiples) and the remainder is divided by

appropriate one’s…………..ten’s………. Here the one’s……ten’s………………..of the later is one step-down

above step because the remainder of initial contains the levels of digits of required plus the digits

below the level so they are removed by integer division. Here level means digit’s place from right

(one’s, ten’s, hundred’s……).

#include<stdio.h>

#include<conio.h>

void main() {

int digit[15],d,num,i=0,j,a=1,b=10,temp;




13

scanf("%d",&num);

temp=num;

while(temp>0) {

d=num%b;

digit[i]=d/a;

a=a*10;

b=b*10;

temp=temp/10;

i=i+1;

}

for(j=i-1;j>-1;j--){

printf("\nThe digit in %d place is %d",j+1,digit[j]);

}

getch();}

Here a variable temp is required to continue loop for number of digits available. A number (decimal)

continuously divided by 10 for n times until quotient 0, gives number of digits because each digit

disappears in integer division by 10 (base of decimal system).

Similarly we can convert the number to string and its each character can be taken by string function

and then its (each character’s) integer value can be reversed back. It is because there are string

functions that work on characters.

Let num be a character digit then

X=num-‘0’→zero

It gives difference in ASCII values of num and zero i.e. actual number (integer value)

Number is converted to character like ‘123’

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

int j;

char num[15];

char dconvert[15];


scanf("%s",num);

strcpy(dconvert,num);

for(j=0;j<strlen(dconvert);j++){

printf("\nThe digit in %d place has ASCII value = %d",j+1,dconvert[j]);

}

getch();}



14

PALINDROME

PALINDROME FOR NUMBER

A palindrome number is that number for which the digits scanned from front is equal to glanced

from back. For e.g. 12321, 121, 125797521. Here we are taking each digit and converting the digits

(of number) to values. Here we are not sure with how many digits are there so this can be solved

especially by two ways- one connecting each numbers(digits) by converting to character and

concatenating them and finally string to integer value and then comparing the value with original

value to find whether it is palindrome or not.

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

void main() {

int j,number=0,digit;

char num[15];

char dconvert[15];


scanf("%s",&num);

strcpy(dconvert,num);

for(j=strlen(dconvert)-1;j>=0;j--){

digit=dconvert[j]-'0';

number=(number*10)+digit;

}

printf("The reversed number is\t%d\n",number);

if(atoi(num)==number)

printf("It is palindrome");

else

printf("It is not Palindrome");

getch();}

In above process after a digit is calculated then the previous number is multiplied by 10 (if available)

so as to increase its level i.e. from one’s to ten’s, ten’s to hundred’s and so on and the digit is added

to it to make required number. As for the case we take second digit from left of 51 i.e. ‘5’ whose

succeeding digit is ‘1’ already stored in ‘number’ is multiplied by 10 and added to new digit, i.e.,

(1*10)+5, that equals 15 which is reverse of 51 but not palindrome number. Similar method is used

in below program.

But for the process using numerical tools, we first up all find digits and each previous digit (if

available) then multiplied by suitable and then the new digit is added to. This process continues till

digits come. Here suitable means 10 each time because ten is the base of decimal or each digit differ

from its left digit by ten’s.



15

Here ‘atoi’ defined in <stdlib.h> converts string to integer.

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

void main() {

int j=0,number=0,digit=0,digita=0,numa=0;

char dconvert[100],num[10];


scanf("%s",&num);

int a=0;

for(j=(strlen(num)-1);j>=0;j--){

digita=num[a]-'0';

a+=1;

numa=(numa*10)+digita;

digit=num[j]-'0';

number=(number*10)+digit;

}

if(numa==number)

printf("\nIt is palindrome");

else

printf("\nIt is not Palindrome");

getch();

}

String can be converted to number directly by function 'atoi' but it can be done taking a character of

string each time and subtracting it to ASCII value of zero i.e. '0'(ASCII values are those particular

numbers representing some characters, letters or any symbol in computer).In these processes we

can take characters of string either from front or from back but the index ‘num[j]’ in above should be

in increasing and decreasing respectively from 0 to length-1 and length-1 to 0 respectively.

AN ANALYSIS OF RETURNING BACK NUMBER FROM DIGITS

The method is to find each digits similar to above and then each digit is multiplied by suitable and

then added to previous digit (or number if available). Here suitable means power of 10 because the

new value means increase in ten’s level.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main() {

int i,num,number,digit,lastdigit=0;


scanf("%d",&num);

for(i=0;num>0;i++) {

digit=num%10;

num=num/10;

int temp=digit*pow(10,i);



16

printf("\n%d is added to%d",temp,lastdigit);

lastdigit=digit*pow(10,i)+lastdigit;

}

printf("\nThe supplied number was%d",lastdigit);

getch();

}

PALINDROME FOR STRING

Similar to palindrome numbers, palindrome strings are those strings in which characters glanced

from front is equal to glanced from back. For e.g. radar, madam, yagay, eye, eve.

1. Using reversing string function

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

char string[10];

char temp[10];

printf("Enter a String");

scanf("%s",string);

strcpy(temp,string);

strrev(string);

if(strcmp(temp,string)==0)

printf("\nThe String is Palindrome");

else

printf("\nThe String is not Palindrome");

getch();}

OR

#include<conio.h>

#include<string.h>

void main() {

char string[100],comp[100];


scanf("%s",string);

strcpy(comp,string);

strrev(string);

if(strcmp(comp,string)==0){


}

else {




17

}

getch();}

Here ‘temp’ variable is for transition case and sudden change in ‘string’ before comparison.

2. A char[] is declared where each time a new last character from string is taken one by

one as array and then checking is done to original string

#include<stdio.h>

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

int i;

char temp[2];

char string[30]="",newstring[30]="";


scanf("%s",string);

for(i=strlen(string)-1;i>=0;i--) {

temp[0]=string[i];

strcat(newstring,temp);

}

if(strcmp(newstring,string)==0){

printf("hello..........");

printf("the String is Palindrome");

}

else {

printf("hello");


}

getch();

return 0;

}

Here the ‘newstring’ is to hold old value which is every time concatenated with new value.

3. Palindrome for string containing even or odd quantity of characters

This method uses the property of even and odd in palindrome that half alphabets taken from

front if equals to half alphabets taken from back then it is palindrome. But for odd (half-1/2) is

compared because the central alphabet is always constant as its pace from front is equal to from

backward. For e.g.

In case of even number of characters containing string like “YAGGAY”, here we need to read first

three characters from ‘Y’ to ‘G’ and store in ‘front’ variable then we need to read last three

characters from ‘Y’ to ‘G’ in reverse manner and store in ‘reverse’ variable. If ‘reverse’ is same to

‘front’ variable then it is palindrome, for the particular case it is palindrome.



18

In case of odd numbers of characters containing string like “YAGAY”, here we need to read first

two characters “YA” and store it in ‘front’ and similarly read last two characters “YA” in reverse

manner and store in ‘reverse’ and comparison is done. Here the central character “G” holds the

same value for the same position when read either from front or from reverse, so the

comparison for this central character is rejected from ‘front’ and ‘reverse’ variables.

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

int i;

char string[100],front[100]="",reverse[100]="";


scanf("%s",string);

if(strlen(string)%2==0){//for case of even

strncat(front,string,strlen(string)/2);

strncat(reverse,strrev(string),strlen(string)/2);

}

else {//for case of odd

strncat(front,string,(strlen(string)-1)/2);

strncat(reverse,strrev(string),(strlen(string)-1)/2);

}

if(strcmp(front,reverse)==0)


else


getch();}

4. Palindrome for string using for-loop

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

int i;

char string[100]="",newstring[100]="";


scanf("%s",string);

static char temp[1];

for(i=(strlen(string)-1);i>=0;i--) {

temp[0]=string[i];

strcat(newstring,temp);

}

if(strcmp(newstring,string)==0){

printf("\nThe String is Palindrome");}

else{




19

}

getch();

return 0;}

ELECRTICITY AUTHORITY

An electricity board charges according to the following rates:

Units Charges (In NRs.)

0-100 40

101-200 50

201-ABOVE 60

All users are also charged ‘meter charge’, which is equal to Rs. 50. Write a complete C program to

read the number of units consumed and print out the total charges.

Here the first condition is to check whether the unit is within range 0 to 100 or 101 to 200 or 201 to

above. After this we calculate the charge for units above higher maximum range then the units is

reduced for lower range continuously whereas on the other hand, charge is added systematically

from higher to lower level. Here ‘goto’ has a special purpose of direct connection. It connects higher

range to lower range before this, units of higher are reduced to lower because ‘goto’ sends to lower

where calculations of lower range only is necessarily to be done.

/*ELECTRICITY AUTHORITY CHARGES AS FOLLOW


0 to100 R.s.40

101 to 200 R.s.50

201to ABOVE R.s.60

AND IT TAKES Rs. 50 as meter charge. A PROGRAM THAT INTAKE UNITS CONSUMED AND FIND OUT

TOTAL CHARGE.*/

#include<stdio.h>

#include<conio.h>

void main() {

int units=0,charge=0,choice=0,totalcharge=0;

printf("ELECTRICITY AUTHORITY CHARGES AS FOLLOW\nUnits----Charges (In NRs.)\n0 to100

R.s.40\n\n101 to 200 R.s.50\n\n201to ABOVE R.s.60\n\nAND IT TAKES Rs. 50 as meter charge\n");

printf("\nEnter Units Consumed\t");

scanf("%d",&units);

int old=units;

if(units>200){

choice=200;

}

else if(units>100){

choice=100;

}

else {

choice=0;

}



20

switch (choice){

case 200:

charge=(units-200)*60;

units=200;

totalcharge=charge;

printf("\nAbove '200' units i.e. 201 to %d (%d * 60)= %d",old,old-200,charge);

old=200;

goto A;

case 100:

A:


totalcharge=totalcharge+charge;

units=100;


old=100;

goto B;

case 0:

B:

charge=units*40;


printf("\nAbove '0' units i.e. 0 to %d (%d * 40) = %d",old,old-0,charge);

totalcharge+=50;

printf("\n Meter Charge = %d",50);

break;

default:

printf("Invalid Input");

}

printf("\n-------------------------------------------------------");

printf("\nThe total cost of power consumed is NRs. %d",totalcharge);

getch();

}

We can improve its continuity by adding a line and a index using ‘goto’ statement, like this.

/*ELECTRICITY AUTHORITY CHARGES AS FOLLOW


0 to100 R.s.40

101 to 200 R.s.50

201to ABOVE R.s.60



21

AND IT TAKES Rs. 50 as meter charge. A PROGRAM THAT INTAKE UNITS CONSUMED AND FIND OUT

TOTAL CHARGE.*/

#include<stdio.h>

#include<conio.h>

void main() {

printf("ELECTRICITY AUTHORITY CHARGES AS FOLLOW\nUnits----Charges (In NRs.)\n0 to100

R.s.40\n\n101 to 200 R.s.50\n\n201to ABOVE R.s.60\n\nAND IT TAKES Rs. 50 as meter charge\n");

int units=0,charge=0,choice=0,totalcharge=0;

maina :

printf("\nEnter Units Consumed\t");

scanf("%d",&units);

int old=units;

if(units>200){

choice=200;

}

else if(units>100){

choice=100;

}

else {

choice=0;

}

switch (choice){

case 200:


units=200;

totalcharge=charge;


old=200;

goto A;

case 100:

A:



units=100;


old=100;

goto B;

case 0:

B:

charge=units*40;


printf("\nAbove '0' units i.e. 0 to %d (%d * 40) = %d",old,old-0,charge);

totalcharge+=50;

printf("\n Meter Charge = %d",50);

break;

default:

printf("Invalid Input");

}

printf("\n-------------------------------------------------------");

printf("\nThe total cost of power consumed is NRs. %d",totalcharge);

printf("\n\n do you continue\n press 1 to continue");



22

if(getch()=='1') {

goto maina;}

}

Here ‘maina’ index repeats the statements only if ‘1’ value is entered. ‘getch()’ takes a character so

‘1’ is user to make 1 as character.

PU2006SPRING

3b) Electricity board charges according to the following rates.

For the first 100 units -------------------------- Rs. 40 per unit

For the next 200 units ------------------------- Rs. 50 per unit

Beyond 300 units ------------------------------ Rs. 60 per unit

All users are also charged ‘meter charge’, which is equal to Rs. 50. Write a complete C program to

read the number of units consumed and print out the total charges.

Next analyzing way in a little bit detail

#include<stdio.h>

#include<conio.h>

/*

units range cost

301 - Rs. 60

101 - 300 Rs. 50

1 - 100 Rs. 40

*/

void main() {

float units=0,cost=0;

printf("\nEnter Units Consumed:\n");

scanf("%f",&units);

if(units>300) {

cost=100*10+200*50;

units=units-300;



23

cost=cost+(units*60);

}

else if (units>100) {

cost=100*40;

units-=100;

cost=cost+(units*50);

}

else {

cost= units*40; // the only last possible condition units>0

}

cost=cost +50;//we can replace it by using this to each blocks of if-elseif ladder

/*In scase instead of 50 as charge if 15% is taken as any kind of sharge like meter charge or vat

then 'cost=115/100*cost' or'cost=cost+(15/100*cost)' can be used*/

printf("The total cost of %f units is %f",units,cost);

getch();

}

--------------------------------------------program termination----------------------------------------------

Next view for analyzing problem in next way but for these no real program codes are written

/*

units range cost

1 - 100 Rs. 40

101 - 300 Rs. 50

301 - Rs. 60

*/

if(units<100) {

cost=units*40;

}

elseif(units<300) {

cost=100*40;

units-=100;

cost=cost+unit*50;

}

else {

cost=100*40+50*200+(units-300)*60;// a single statement like this can also work

}

cost+=50;

---------------------------termination of one way of viewing program------------------------------

Next view for analyzing problem in next way but for these no real program codes are written

(by simple if statements using logical operators)

declare global_units=units;

if(units<100 { // this can ber replaced bu units>0 && units<100

cost=units*40;



24

}

if(units<300 && units>100) {

cost=10*/40+(units-100)*50;

}

if(units>300) {

cost=100*40+200-50+(units-300)*60;

}


The coming process is similar to done in ELECRTICITY AUTHORITY already but this process is

simple in the manner that it doesn't use ‘goto’, which may be confusing for readers to analyze.

This coming method uses the technique of dividing operations in parts in such a way that each

part is executed if only the condition in 'if' is satisfied. If the condition is satisfied then some

operations is done. This operation is related to the condition of the particular 'if' then the

necessary quantity whose calculation is done is removed from the value to which calculation is

to be done as a whole. Realizing problem in codes speak more than words or programming

analysis is great than language taking programming.

if(units>0) {

if(units>100){

cost=100*40;

units-=100;

}

else {

cost=units*40;

units=0;

}

}

if(units>0) {

if(units>200){

cost=cost+200*50;

units-=200;

}

else {

cost=cost+units*50;

units=0;

}

}

if(units>0) {

cost=cost+units*60;

units=0;

}


Here block of ‘if-statement’ checks condition as follow:



25

�� block �� part Checks units >100 �� block �� part Checks units <100 �� block �� part Checks units >200 �� block �� part Checks units <200 �� block �� part Checks units >300

Here each block has specified work to do. ��Block calculates the sum for the units between 1 to 100

units and then it extracts the calculated portion from the total consumed units by subtracting units

to calculated amount of units. So that the next block of ‘if’ needs to calculate the sum for next range

of units, i.e. 2of 10 units to 300 units but its range is 200 and as calculated part (1 to 100) is

extracted so

------------0-------------------------100--------------------200--------------------300--------------

It is extracted It is left

In the last part, we similarly do for the range (101-300) and finally the left all is done by �� block.

Here in each terminating of �� part of each block, units is assigned zero. Sothere is no necessary

of further calculations. It is because the �� part of each block (except ��block) works taking the

condition on the basis that it works for particular range with leaving some range for succeeding

block. It means, for e.g. �� part of ��block works in such that in it works for particular range of

101-300 and leaving 300-above range for ��block so here units is still left to be calculated for ��block and this all is the reflection of the condition of if of ��part of ��block if (units>200) i.e.

from 101 to 300. For e.g.

If 500 units is consumed then calculation takes as below:-

(Termination in �� block)

Block-��

units>0 500>0 True - - -

Block-��

units>100 500>100 True cost=100*40

units-=100

4000

500-100

4000

400

Block-��

units>0 400>0 True - - -

Block-��

units>200 400>0 True cost=cost+200*50

units-=200

4000+200*50

400-200

14000

200

Block-��

units>0 200>0 True cost=cost+(units*60)

units=0

14000+200*60 26000



26

Let us assume consider for value units 60 that must terminate in ��block

Block-�� units>0 60>0 True - - -

Block-�� units>100 60>100 False cost=units*40

units=0

60*40

0

2400

0

Block-�� units>0 0>0 False - - -

Block-�� units>0 0>0 False - - -

Let us consider for value units 150 that terminate in �� block:-

Block-��

units>0 150>0 True - - -

Block-��

units>100 150>100 True cost=100*40

units-=100

100*40

150-100

4000

50

Block-��

units>0 50>0 True - - -

Block-��

units>0 50>200 False cost=cost+(unit*50)

unit=0

4000+(50*50)

0

6500

Block-��

units>0 0>0 False - - -

Actually in designing problem, we need to consider ourself as programmer and we need to analyze

all the steps or execution that takes for each possible input of users. Then we must realize in writing

and executing. Here in above program, we need to analyze the execution of value of cost and units

for each possible input of users.

There are many possible conditions even including simple ‘if’, it depends on flow of direction of

executing from top range to low range or vice-versa.



27

ARRAY

The derived structured data types are more effective in C because user can himself declare according

to his requirement where every element has some relationship forming a structure. Arrays are

important regarding structure type data and these structures can be efficiently, easily used on user-

interference dynamically by numbers, characters or strings (sum of characters). Names and locations

are unique for the ordinary purposes but when we want to use a common term related to both we

use array. Arrays are two-one that is fixed i.e. static whose declaration is completed before

initialization or running of main part of program. Such array are less effective, easier and the only

two purposes of such array is to-one for just demonstration of way of using array or how array works

and the other may be making collection of characters or numbers in small space rather than making

separate space for each terms. The other dynamic is vast because it include static plus the working

that is not possible from any else (even static). The dynamic array works on the principle of two-one

in which its limit is fixed by program and the other is made by user himself. The later one is more

effective. For this the program either directly or indirectly or through process of calculations declares

the size of array. To make array’s cycle, we use loop with a variable different each time (by+1, -1 or

any according to the conditions). Usually we declare by incrementing each value by 1 because in

array numbers are taken in sequential addresses starting from 0 to one number less than total

quantity of elements. For this purpose we can use, if and using counter and goto, similarly while,

counter or simply for. The easier one, concise one is for, we generally use it except for case where if

is used when condition is needed more to be focused, goto if direct linkage is necessary immediately

or in some making program more effective, fast etc. However using if and exit inside for-loop can

work. In array, we use for-loop generally from (i=1 to n) to take numbers and give them to addresses

i=0 to i=n-1 because in array address starts from 0 but quantity is n, so last address is n-1.

Similarly, we use such loop for output of numbers or characters. It they are sequentially to be

printed out we use i=0 to n-1 (i=1 to n) incrementing or decrementing (i=n to 1 or i=n-1 to 0) by 1 but

sometimes a gap of addresses is required for this appropriate increment or decrement is done. For

the case of sorting-bubble, selection, two loops were taken and array calculation was done. This is

illustrated in Loop Section. However while understanding array calculation we must be clear about

loop. Because it is only way to read or write the array elements in its proper address of finite or

infinite array elements (infinite here means in the way that we don’t know the exact quantity of

elements as it is user-dependent). Any program must be logically and technically strong enough to

hand no limitations regarding quantity and quality of values provided by users for unlimited time

purpose.

#include<stdio.h>

#include<conio.h>

void main() {

int number[5],i;

for(i=0;i<5;i++) {


scanf("%d",&number[i]);

}

for(i=0;i<5;i++) {

printf("\nThe %d th number is %d",i+1,number[i]);

}

getch();

}



28

Besides reading and writing, loops are used to continue many cycles in arrays, some for

comparisons, some for arithmetic and logical calculations, some for arrangements and many more.

In sort, there were two loops for comparison and interchanging calculations. Look sorting for detail

analysis. However in this we will be concerned with array only.

In each loop, we find different sequentially coming array elements compared with other

sequentially. In this way every array elements got compared and our purpose satisfied. Such loops

and arrays are just general purposes. We can extend it to 2-D, 3-D or even multi-dimensional

purpose. Each time we use more dimension, we need a looping more i.e. if 2 loops for 1-D then for

the same purpose 3 loops is necessary for 2-D and similarly for others because one more loop is

necessary for 1-D to 2-D because in each step of D, there is one more necessity of a loop in defining

array elements. For 3-D process, for same one more loop is required to find the one more dimension

of array. So regarding the purpose of sorting, 2 loops for 1-D, 3 loops for 2-D, four loops for 3-D and

so on for multi-dimensional.

2D:

In 2-d array, two co-ordinates are use to specify the point of address as there are rows and columns.

So the two loops to define 2-D array elements.

#include<stdio.h>

#include<conio.h>

void main() {

int number[2][2],i,j;

for(i=0;i<2;i++) {

for(j=0;j<2;j++) {


scanf("%d",&number[i][j]);

}

}

for(i=0;i<2;i++) {

for(j=0;j<2;j++) {

printf("\nThe number in %d row and %d column is \t %d",i+1,j+1,number[i][j]);

}

}

getch();

}



29

a b result

1 1 to 1 1

2 1 to 2 12

3 1 to 3 123

4 1 to 4 1234

PU2008FALL 2b)

WAP that asks a user for a number and finds out if the number is present in the D array given

below:

int arr[3][3] ={{6,37,33},

{12,11,13}'

{14,85,96}

};

#include<stdio.h>

#include<conio.h>

void main() {

int found=0;

int i,j,num,arr[3][3]={ {6,37,33},

{12,11,13},

{14,85,96}

};

printf("Enter a number:\t");

scanf("%d",&num);

for(i=0;i<3;i++){

for(j=0;j<3;j++){

if(arr[i][j]==num) found++;

}

}

if(found==1)printf("Found");

else printf("Not Found");

getch();

}

In this program, it uses one of the techniques of declaring values in 2D-array. Here row elements are

6, 37and 33, row elements are 12, 11and 13 and row elements are 14, 85, and 96.

Row

Column

0 1 2

0 6 12 14

1 37 11 85

2 33 13 96

GREATEST AND SMALLEST NUMBER AMONG ‘N’ NUMBERS (using array)

Here a variable greatest and smallest is declared and then the comparison is done with the variable

to other numbers serially coming in arrays by the help of for-loop. According to our necessity-



30

greatest or lowest, the sign of comparison is assigned and value is assigned if condition is true. After

the completion of all comparison, we get greatest and smallest among n numbers.

#include<stdio.h>

#include<conio.h>

void main() {

int i=0,j=0,number,num[100],great,small;

printf("\n Enter how many numbers?\n");


printf("\n\n\n Enter numbers\n");

for(i=0;i<number;i++) {

scanf("%d",&num[i]);

great=small=num[0];

if(num[i]>great) {

great=num[i];

}

if(num[i]<small){

small=num[i];

}

}

printf("The greatest number is %d\t and the smallest number is %d",great,small);

getch();

}



31

PATTERNS OF CHARACTERS AND NUMBERS

For the purpose array is necessary because we deal with pointer rather than values even though it

may seem that the output has connection with values. [NOTE: In some values have connections for

this looping may be sufficient.] After talking about arrays, generally looping comes itself for the

general case of dynamics. The output of Floyd’s triangle is a 2-D. so two loops are sufficient (in

simple cases), one outer loop defines the rows and the second inner loop is for printing the pattern

of columns corresponding to each rows declared by outer loop. The main analysis is needed for inner

loop because the outer loop is just for counting the rows but inner has many things to do. The inner

lop must develop a formula connecting the outer loop’s value with the patters to be printed and its

connection to further coming patterns of inner loop. For this purpose, counter may be declared,

incremented, decremented, multiplied, divided or reduced (in case of strings) or many variables that

may be transitional or temporary or others can be declared according to the availability of resources

and values of outer loop for the purpose of pattern resembling columns of corresponding outer

loop. The above all we discussed is all about loops and arrays however our pattern can also be

achieved by scanf and printf. These formatted functions use various field widths for various outputs

but we need to develop a formula analyzing the printing patterns with the field width. These

functions generally work for characters, however it works more conveniently with numbers but

numbers being arithmetic are implemented by arithmetic process rather than by functions.

However, output doesn’t vary. In some, it is necessary for using such functions. Some of the tips for

using these functions are as below:-

1. The field width is less than length of string implies that all to be printed as usual.

2. The value after decimal (if is present) implies the number of characters to be printed. If it is

zero, nothing is printed.

3. The value before decimal makes the space to which printing is done. The printing is done in

such a way from right (in reference to space) but from left (in reference to values it have).

4. The negative sign causes no gap in printing i.e. top-most left shifting.

5. If nothing is before decimal then the printing is usual however number after decimal

restricts (if present).

An example to demonstrate the above function

#include<stdio.h>

#include<conio.h>

void main(){

int i;

char string[]="NEPAL";

for(i=1;i<=5;i++) {

printf("%-5.*s\n",i,string);

}

getch();

}

Output:

N

NE

NEP

NEPA

NEPAL



32

Here in the function * is the value supplied by i. this function format is usually used in UNIX.

Similarly, we can replace * by our necessary number and for this a more number or variable is

declared in argument part. Analysis:- In the program, ‘5’ defines the space to which printing is done

it is every time (or for every loop it is 5), the negative sign shifts the output to left-most part. The ‘*’

is pointed to I, whose value change from 1 to 5 so that each time the loop continues, the number of

patterns increases serially one by one as the value of I of loop increases. So even though the output

is in two-dimensional but a single loop works. Because the loop does two ting in this case (this case

is for using functions), one it provides the rows to be printed and other it defines the number of

characters to be printed corresponding to the row. For the case the number of rows is equal to the

number of characters to be printed, so there is no necessity of declaring, calculating any more

variables or any.

SCNAF FUNCTION FOR NUMBERS:-

All the above five tips are similar with two new things- one is 0 is declared before some number

value before decimal point, then the spaces vacant are filled by zeros i.e.

printf("%07d",9855) →

And other is that d can be replaced by ld, hd, d for long integers, short integers, etc. Here number

before decimal gives total space and number after decimal gives space for decimal values. But the

case for exponential is little different that part, that exponential takes the values according to the

number value available and take separate place whose place can’t be defined.

For e.g.

float y=223.34568;

printf("%-10.2e",y); →

Note: printf(“%*.*f”,width,precision,number) is valid.

These functions are used for various patterns but especially these are used to concise large value in

terms of exponent with proper width and precisions.

PATTERNS OF CHARACTERS AND NUMBERS USING PRINTF FUNCTIONS:-

For this case, as our patterns are in 2-D so usually two loops are required, however in some cases

one loop works (if the value of one loop can compensate other also). One loop is for outer purpose

of row and other is for column purpose.

Similar to the 2-D array, here the purpose of two loops is similar. The main focus is on inner loop

because the pattern is determined by it.

0 0 0 9 8 5 5

2 . 2 3 e + 0 0 2



33

BASIC PRINCIPLE OF PRINTING PATTERNS

Here two loops need to be required as general case but (being a special case) here) the row equals

to the number of characters of the string “NEPAL” to be printed. So here one loop is efficient if

concatenating function is used.

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main(){

int i;

char a[]="NEPAL";

for(i=0;i<5;i++) {

char final[5]="";

strncat(final,a,i+1);

printf("%d%s",i,final);

printf("\n");

}

getch();

}

Output:

0N

1NE

2NEP

3NEPA

4NEPAL

#include<stdio.h>

#include<conio.h>

void main(){

char a[]="NEPAL";

int i,j;

for(i=1;i<=5;i++) {

for(j=1;j<=i;j++) {

printf("%c",a[j-1]);

}

printf("\n");

}

getch();

}

Output:

N

NE

NEP

NEPA

NEPAL



34

As usual it uses two loops. Here the second is adjusted in such a way that it takes values from 1 to

number of rows it is present or from 1 to the value of outer loop it corresponds to. This is the basic

principle of making patterns. The above starts from 1 to 5, increasing 1 at a time means increasing 1

at time for each row increasing.

#include<stdio.h>

#include<conio.h>

void main(){

char a[]="NEPAL";

int i,j,k;

for(i=1;i<=5;i++) {

for(k=4;k>=i;k--){

printf(" ");

}

for(j=1;j<=i;j++) {


}

printf("\n");

}

getch();

}

Output:

N

NE

NEP

NEPA

NEPAL

Here as the rows increases, space decreases and in the same wave number of characters increases.

However ‘k’ is essential to declare that gives required space because here the value of k is

decreasing but I and j are increasing.

From above we can make a conclusion that the second loop is important and the main base is the

second loop which under takes the consideration of outer-loop.

We did for strings and similar is for numbers with the same basic fundamental principle.

Note: There are generally two ways of patterns- one is using string function directly and the

other is using character one at each time by 2d-loops like ‘for-loop.’ We have practiced both

the methods in the examples above.



35

SOME MORE PATTERNS

PU 2005 FALL5

IF INPUT IS “NEPAL” then output must be

1N

2EE

3PPP

4AAAA

5LLLLL

This program shows how the value of ‘i’ can be used in both- characters printing and number

printing. The corresponding value of ‘i’ for number printing is one more than for characters as

sharacter array starts from zero.

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main(){

char a[50];

int i=0,j=0;

printf("Enter a string");

scanf("%s",&a);

for(i=0;i<strlen(a);i++) {

printf("%d",i+1);

for(j=0;j<=i;j++) {

printf("%c",a[i]);

}

printf("\n");

}

getch();

}

Nepal

Nepa

Nep

Ne

N

#include<stdio.h>



36

#include<conio.h>

int main(){

char a[100]="Nepal";

int i=0,j=0,k=0;

for(i=1;i<=5;i++) {

for(j=1;j<=(5-i+1);j++) {


}

printf("\n");

}

getch();

return 0;

}

Hello

Hello

Hel

He

H

#include<stdio.h>

#include<conio.h>

int main(){

char a[100]="hello";

int i=0,j=0,k=0;

for(i=1;i<=5;i++) {

for(k=1;k<i;k++){

printf(" ");}

for(j=1;j<=(5-i+1);j++) {


}

printf("\n");

}

getch();

return 0;

}

h

he



37

hel

hell

hello

#include<stdio.h>

#include<conio.h>

int main(){

char a[100]="hello";

int i=0,j=0,k=0;

for(i=1;i<=5;i++) {

for(k=1;k<i;k++){

printf(" ");}

for(j=1;j<=i;j++) {


}

printf("\n");

}

getch();

return 0;

}

1

12

123

1234

12345

#include<stdio.h>

#include<conio.h>

int main(){

int number;



int i=0,j=0,k=0;

for(i=1;i<=number;i++) {

for(k=1;k<i;k++){ //space management

printf(" ");}

for(j=1;j<=i;j++) { //printing number from 1 to i for each line given by 'i'

printf("%d",j);



38

} printf("\n");}

getch();

return 0;

}

�� = � ��!

�

��

�� = 1 + + ��! +

��! + ��

�! + ��! + ��

�! +…………..� �!

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main(){

printf("Enter power of e");

int x;

scanf("%d",&x);

printf("Enter n to be run");

int n;

scanf("%d",&n);

int i,j;

float sum=0;

float fact=1;

for(i=0;i<=n;i++){

float value= pow(x,i);

for(j=1;j<i;j++) fact=fact*j; /*factorial calculation*/

sum=sum+(value/fact);

}

printf("sum is %f ",sum);

getch();

return 0;

}

!"# = 1 � − 2 � + 3 � − 4 � + 5 � − 6 � + ⋯ … … , ./#!

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main(){

printf("Enter value of x");

int x;

scanf("%d",&x);




39

int n;

scanf("%d",&n);

int i,j;

float sum=0;

for(i=1;i<=n;i++){

float value= pow(x,i);

if(i%2==0) {

sum=sum-i/value; /*for the case of even index sign is negative*/

}

else

{

sum=sum+i/value; /*for the case of odd index sign is positive*/

}

}


getch();

return 0;

}

PU2004

!"# = 01! − �

2! + �3! − �

4! + �5! . . … … … , ./#!

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main(){

printf("Enter value of x");

int x;

scanf("%d",&x);


int n;

scanf("%d",&n);

int i;

float sum=0,term;

term= pow(x,1);

for(i=1;i<=n;i++){

sum=sum+((-x*term)/i);

}


getch();

return 0;

}

Here we are using a pattern .�=� 23∗�5�

�



40

Where .� represents ,�6 term and n is given by the for-loop; by i which gives the position of the

term and x is the input given by user. We can do this by next method using a function that finds out factorial and taking help of i to find

power of x and changing sign +,- after each term is found out and adding to sum , which is printed at

last.

Pu2011-fall-4b

WAP to print the series upto term is less than 250

Series

0�7�� ,

��7�� ,

��7�� ,

��7�� ,

��7��8 , ….…… upto term>250

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main(){

int i=1;

float sum=(pow(1,2)+pow(2,2))/3, oldsum=0;

while(sum<250) {

i+=1;

oldsum=sum;

sum=(pow(i,2)+pow(i+1,2))/(i+2);

}

printf("sum less than 250 is %f",oldsum);

getch();

}

Here actually at first the first term is kept in sum in initialization then it is checked and printed then i

is incremented to meet the next term of the series but before printing it is checked strictly. Here

sum is kept in float because we are not sure that the term is in int or not.

1111

1111

1111

1111

#include<stdio.h>

#include<conio.h>

int main(){

int i=0;

for(i=1;i<=16;i++) {

printf("%d",1);

if(i%4==0){ printf("\n");}

}



41

getch();

return 0; }

Here we are using a single loop of i, as after the numeric value of i divisible (4, 8, 12, 16) is printed

then line is broken out.

1

12

123

1234

#include<stdio.h>

#include<conio.h>

int main(){

int i=0,j=0,k=0;

for(i=1;i<=4;i++) {

for(k=1;k<=(4-i);k++){ /*to maintain required spacing*/

printf(" ");

}

for(j=1;j<=i;j++) {

printf("%d",j);

}

printf("\n");

}

getch();

return 0;

}

hints

1 1�

121 11�

12321 111�

1234321 1111�

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main(){

int i=0,j=0,k=0;

int num=0;

int counter=0;

for(i=1;i<=4;i++) {

counter=counter*10+1;

for(k=1;k<=(4-i);k++){

printf(" ");



42

}

num=counter*counter;

printf("%d",num);

printf("\n");

}

getch();

return 0;

}

1

12

123

1234

#include<stdio.h>

#include<conio.h>

int main(){

int i=0,k=0; for(i=1;i<=4;i++) {

for(k=1;k<=i;k++){

printf("%d",k);

}

printf("\n");

}

getch();

return 0;

}

4321

432

43

4

#include<stdio.h>

#include<conio.h>

int main(){

int i=0,k=0;

for(i=4;i>=1;i--) {

for(k=4;k>=(4-i+1);k--){

printf("%d",k);

}

printf("\n");



43

}

getch();

return 0;

}

0.1 0.03 0.005 0.0007 0.00009 ……. for n terms

//0.1 0.03 0.005 0.0007 0.00009 ……. for n terms

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main(){

int n;

double counter=1;

printf("enter value of n");

scanf("%d",&n);

int i;

int zeroafterdecimal=0;

for(i=1;i<=n;i++) {

counter/=10;

zeroafterdecimal++;

printf("%f\t",counter);

counter=counter+(2/pow(10,zeroafterdecimal));

}

getch();

return 0;

}

Floyd’s triangle

1

10

101

1010

10101

101010

.

.

.

upto n lines

/*1



44

10

101

1010

10101

101010

*/

#include<stdio.h>

#include<conio.h>

int main(){

int n;


scanf("%d",&n);

int i,j;

for(i=1;i<=n;i++) {

for(j=1;j<=i;j++) {

if(j%2==0) printf("0");

else printf("1");

}

printf("\n");

}

getch();

return 0;

}

5 55 555 5555 55555…………………………. upto n terms

//5 55 555 5555 55555….. upto n terms

#include<stdio.h>

#include<conio.h>

int main(){

int n;


scanf("%d",&n);

int i,j;

int count=5;

for(i=1;i<=n;i++) {

printf("%d\t",count);

count=count*10;

count+=5;

}

getch();

return 0;

}

NEPAL

EPA

P



45

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

printf("enter a string (odd length)\t");

char string[40];

scanf("%s",&string);

int length=strlen(string);

int i,j,k;

int tim=length;

for(i=1;i<=((length/2)+1);i++){

for(k=1;k<i;k++)

printf(" ");

for(j=i;j<=tim;j++){

printf("%c",string[j-1]);

}

tim=tim-1;

printf("\n");

}

getch();

return 0;

}

OR

/*

NEPAL

EPA

P

*/

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

printf("enter a string (odd length)\t");

char string[40];

scanf("%s",&string);

int length=strlen(string);

int i,j,k,con=0;

for(i=length-1;i>=0;i--){

for(k=i;k<length-1;k++)

printf(" ");

for(j=con;j<=i;j++){

printf("%c",string[j]);

}

con++ ;

printf("\n");

}

getch();

return 0;

}



46

Here in second case I = 4to 0 shows last position of character of each line up to which j’s value starts

from k=0 in first and then k++ i.e. k=1…..to k=2.however for k=3, i=1 so , j<=i doesn’t satisfy and runs

without any output. However the loop of I runs 2 more then necessary this can be controlled by

adding a condition that is

if (k=i) break;

in this condition the value of j is controlled up less or equal to equal or less.

1

2 2

3 3 3

4 4 4 4

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

int n;

printf("enter a number\t");

char string[40];

scanf("%d",&n);

int i,j,k;

for(i=0;i<n;i++){

for(k=1;k<n-i;k++)

printf(" ");

for(j=0;j<=i;j++){

printf("%d",i+1);

}

printf("\n");

}

getch();

return 0;

}

1

22

333

4444

55555

#include<stdio.h>



47

#include<conio.h>

#include<string.h>

int main() {

int n;

printf("enter a number\t");

char string[40];

scanf("%d",&n);

int i,j,k;

for(i=0;i<n;i++){

for(k=0;k<i;k++)

printf(" ");

for(j=0;j<=i;j++){

printf("%d",i+1);

}

printf("\n");

}

getch();

return 0;

}

PATTERNS OF ‘*’

*

**

***

****

#include<stdio.h>

#include<conio.h>

int main(){

int n;


scanf("%d",&n);

int i,j;

for(i=1;i<=n;i++) {

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();

return 0;

}

****

***



48

**

*

#include<stdio.h>

#include<conio.h>

int main(){

int n;


scanf("%d",&n);

int i,j;

for(i=n;i>=1;i--) {

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();

return 0;

}

*****

****

***

**

*

#include<stdio.h>

#include<conio.h>

int main(){

int n;

printf("enter value of lines");

scanf("%d",&n);

int i,j,k;

for(i=n;i>=1;i--) {

for(k=1;k<=(n-i);k++) {

printf(" ");

}

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}



49

getch();

return 0;

}

*

***

*****

/*

*

***

*****

*/

#include<stdio.h>

#include<conio.h>

int main(){

int n;

printf("enter no of * but odd only else it change to lowest nearest odd");

scanf("%d",&n);

int i,j,k;

int count=0;

for(i=1;i<=n;i+=2) {

count++;

for(k=1;k<=(n-count);k++) {

printf(" ");

}

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();

return 0;

}

*****

***

*

/*

*****

***



50

*

*/

#include<stdio.h>

#include<conio.h>

int main(){

int n;

printf("enter no of * but odd only ");

scanf("%d",&n);

int i,j,k;

int count=0;

for(i=n;i>=1;i-=2) {

count++;

for(k=1;k<count;k++) {

printf(" ");

}

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();

return 0;

}

*

**

***

****

#include<stdio.h>

#include<conio.h>

int main(){

int n;

printf("enter no of lines");

scanf("%d",&n);

int i,j,k;

int count=0;

for(i=1;i<=n;i++) {

count++;

for(k=i;k<n;k++) {

printf(" ");

}

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();



51

return 0;

}

*

**

***

****

Here one space increase by each new line.

#include<stdio.h>

#include<conio.h>

int main(){

int n;

printf("enter no of lines");

scanf("%d",&n);

int i,j,k;

for(i=1;i<=n;i++) {

for(k=0;k<i-1;k++) {

printf(" ");

}

for(j=1;j<=i;j++) {

printf("*");

}

printf("\n");

}

getch();

return 0;

}

SOME MORE DIFFICULT PATTERNS

Pyramid of numbers:

1



52

2 3 2

3 4 5 4 3

4 5 6 7 6 5 4

5 6 7 8 9 8 7 6 5

#include<stdio.h>

#include<conio.h>

void main() {

int i,j,n=5;

for(i=1;i<=n;i++){

for(j=1;j<=n-i;j++)

printf("\t");

for(j=i;j<=2*i-1;j++) printf("\t%d",j);

for(j=2*i-2;j>=i;j--) printf("\t%d",j);

printf("\n");

}

getch();}

Reverse pyramid of numbers

5 6 7 8 9 8 7 6 5

4 5 6 7 6 5 4

3 4 5 4 3

2 3 2

1

#include<stdio.h>

#include<conio.h>

void main() {

int i,j,n=5;

for(i=0;i<=n;i++){

for(j=n-i;j<=2*n-2*i-1;j++)

printf("\t%d",j);

for(j=2*n-2*i-2;j>=n-i;j--) printf("\t%d",j);

printf("\n");

for(j=0;j<=i;j++) printf("\t");

}

getch();}



53

Herein pyramid of numbers and its reverse, we are printing pattern in installments i.e. is in two

parts; one part contains first 5 rows and the second contains left 4 rows. Such patterns are little bit

difficult and are useless for exam-oriented so its details are exclude.

1

01

101

0101

10101

010101

1010101

01010101

101010101

Here ‘back’ variable is used to alter the stored value (it is printed); ‘1’ and ‘0’, ‘back’ variable works

within a line only i.e. it changes value for appropriate values to be printed on same line. In contrast,

‘fback’ variable stores the first character (here ‘1’, ‘0’ are printed as characters) for each line. To

maintain pattern ‘back’ and ‘fback’ are initialized as zero. It is important to note the pattern that old

value of ‘fback’ not only changes ‘fback’ but also ‘back’, but this alternating takes place one time for

each new line, whereas the value of ‘back’ alters within a same line for the number of times equal to

the number of characters to be printed in the particular line. Study the scope of different ‘for-loop’

for this purpose.

#include<stdio.h>

#include<conio.h>

void main(){

int i,j,fback=1,back=1,n=20;

for(i=0;i<n;i++) {

for(j=0;j<=i;j++)

if(back==1) back=0,printf("1");

else back=1,printf("0");

printf("\n");

if(fback==1) fback=0,back=0;

else fback=1,back=1;

}

getch();}



54

1

2 3 2

3 4 5 4 3

4 5 6 7 6 5 4

5 6 7 8 9 8 7 6 5

Here input is “123456789”.

Read the relationship between outer ‘i’ loop and inner ‘j’ loop. Realizing the problem in 2D method,

the rows of the above pattern or lines of above patterns are represented by ‘i-loop’ and similarly the

columns of the above pattern or values at specific positions are represented by ‘j-loop’. These things

are mostly common in all 2D patterns; however the above pattern has many things to state. We

have used a separate variable-‘s’ to manage spacing but we can use ‘i’ variable (comparing ‘i’ with a

constant value) to manage space (such cases were done in former cases). The value of ‘i-loop’ is not

only used in specifying line but also in ‘j-loop’ to specify characters. It is the coincidence that the line

represented by ‘i’ resembles first value to be printed for the line so ‘i’ is used in ‘j-loop’. Similarly ‘k-

loop’ (used to print the reverse mannered numbers) used the property of ‘j’ as only for initialization

purpose and uses variable- ‘a’ but in decremented order to maintain reversed pattern.

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main() {

int i,s,j,k;

char string[100];

printf("Enter a string");

scanf("%s",string);

printf("\n");

if(strlen(string)%2==0) {/*odd number of characters*/

printf("ODD quantity of string is only preferrable");/*this is the limitation of the pattern*/

exit(0);

}

for(i=0;i<((strlen(string)/2)+1);i++) {

for(s=0;s<((strlen(string)/2)-i);s++) {/*space management*/

printf(" ");

}

for(j=i;j<((2*i)+1);j++) {

printf("%c",string[j]);

}

int a=j-2;

for(k=j+1;k<(j+1+i);k++){

printf("%c",string[a]);

a--;

}

printf("\n");

}



55

getch();

return 0;

}

Pascal’s triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Here input is ‘7’.

This pattern is same to binomial coefficients of Newton’s expansion.

The conclusion of the above pattern is

element [row] [column] = element [row-1] [column-1] + element [row-1] [column]

And a necessary vision is that the numbers glanced at from front or from behind each line is ‘1’ for

all the time.

The expression if(j ==0 || j==i){ line[i][j]=1; declares it. Here ‘j=0’ means first value for

each new line and ‘j=i’ is the last possible value of ‘j’ for the particulat line, the other thing we find is

that the value of ‘i’ for not only supports the row of pattern (or number of particular line) but also it

supports the total quantity of numbers to be printed; if(j ==0 || j==i) here ‘i’ has the

second purpose but line[i][j]=1 has the first purpose.

line[i][j]=line[i-1][j-1]+line[i-1][j]; this expresses the expression of above tabulation.

printf("\t\t"); here double tab maintains the necessary pattern between the numbers within a

same line.

#include<stdio.h>

#include<conio.h>

void main()

{



56

int i,j,k,n,line[50][50];

//clrscr(1);


printf("\t");

scanf("%d",&n);

for(i=0;i<n;i++){

for(k=0;k<n-(i+1);k++){/*for tab space management before numbers*/

printf("\t");

}

for(j=0;j<=i;j++){

if(j ==0 || j==i){

line[i][j]=1;

printf("%d",line[i][j]);

}

else {

line[i][j]=line[i-1][j-1]+line[i-1][j];

printf("%d",line[i][j]);

}

printf("\t\t"); /*for tab space management between numbers*/

}

printf("\n\n"); /*for making more clearity of pattern*/

}

getch();

}

PU2002 FALL 5

WAP that takes angle in degrees and give out value of sine for that angle without using ‘sin()’

function of math library.

Actually, do you how your computer or a pocket calculator finds the value of sine? It is the

application of Taylor’s series for sine as follow:

sin x =930! − 9�

�! + 9��! − 9:

8! + 9;<! − … … … … … … … … … ..

here, ‘x’ is in radians. We can convert degrees to radians by : radian==>?@>>∗A

0B�

#include<stdio.h>

#include<math.h>

#include<conio.h>

float factorial(x){

float fact=1; int i;

for(i=1;i<=x;i++){

fact=fact*i;

}

return (fact);

}

void main () {

float degree,radian,sum=0,term;



57

int i,power;

printf("Enter the angle in terms of degrees:\t");

scanf("%f",&degree);

radian=(degree*3.14)/180;

for(i=1;i<=10;i++) {

power=(2*i)-1;

if(i%2==1) {

term=pow(radian,power)/(factorial(power));

}

else {

term=(-1*pow(radian,power)/(factorial(power)));

}

sum=sum+term;

}

printf("Sine(%f) =\t%f",degree,sum);

getch();

}

Here we use factorial that calculates factorial for the term required.

Initially after taking angle in degree we convert it into radians.

Then a for-loop running ten times is used so that we can calculate the sum for ten terms, but this can

be increased to any number (may be 99 or 999) so that we get our required accuracy in float values

after decimal.

Here we are using a pattern which shows that:

1. Each term has power like 1, 3, 4, 5, 7……. i.e. (2n-1), where ‘n’ shows the index of term

indicating the position. For first position n=1 gives power=(2*1-1)=1, similarly for second

term, power(2*2-1)=3, third has (3*2-1)=5 and so on….. So we are using variable ‘power’ to

represent the power for the term.

2. Each term is alternatively positive and negative. First term one is positive, second term is

negative, third term is again positive, fourth term is negative and so on…….. From this

pattern, we see that if the index of term representing the position of the term is odd then

the term is positive else it is negative.

if (i%2==1) THEN term=(pow(radian,power)/(factorial(index)));

else term=(-1*pow(radian,power)/(factorial(index)));

As we know ‘i’ here represents the index so we are checking if for ‘i’ to find whether term is positive

or negative and the statement following calculates required term with appropriate sign ‘+’ or ‘-’.

NEXT METHOD:

#include<stdio.h>

#include<math.h>

#include<conio.h>

void main () {

float degree,radian,sum=0,term;



58

int i;

printf("Enter the angle in terms of degrees:\t");

scanf("%f",&degree);

radian=(degree*3.14)/180;

term=pow(radian,1);

for(i=2;i<=2;i++) {

term=term-1*(term*pow(radian,2))/((2*i-2)*(2*i-1));

sum=sum+term;

}

printf("Sine (%f) =\t%f",degree,sum);

getch();

}

Here we have little bit analyze the pattern of the terms with reference to two things: one is term

preceding the term to be calculated and the other is with the index representing the position of the

term to be calculated. If we study the pattern, we come with many conclusions, one of them may be

like this (this one is used in the program):

1. In the numerator of each term, each term succeeding is multiplied by - to preceding term.

2. In the denominator, for eg. look 3! (of second term) is converted to 5! (of third term), by

multiplying 3! by 4 and 5 one after another consecutively. We may analyze that each new

term is obtained by dividing each preceding term to (2n-1) and (2n-2),i.e.

denominator of any term=denominator of preceding term*(2n-1)*(2n-2), where ‘n’ is the

index representing the term to be found out.

For e.g.

denominator of third term=denominator of second term*[(2*3-1)*(2*3-2)=5*4]. This is how

3! is changed to 5!.

(2n-1)*(2n-2) can be changed to (2n-2)*(2n-1) in order to make the order of ascending

according to our easiness for the pattern of the program.

Similarly, we may change this ‘(2n-2)*(2n-1)’ = ‘2(n-1)*(2(n-1)+1)’ and we may use this in our

code instead of ‘(2n-2)*(2n-1)’, where ‘i’ is used instead of ‘n’ in code.

By above, we come to conclusion,

termG = termG50 ∗ x��2n − 2 ∗ �2n − 1

OR

termG = termG50 ∗ x�2�n − 1 ∗ �2�n − 1

Where, ‘n’ represents the index of the term to be found out.

NOTE: if we can’t any library function then in above instead of ‘pow()’ use the method described

to calculate power without any function of “without function”-section.



59

Conversion

Decimal to binary

2 12

2 6 0

2 3 0

2 1 1

0 1

So �120� = �1100�

Here ‘factor’-variable works as a multiplier to obtain appropriate numbers of appropriate digits,

please refer ‘Digits taking from number’ for further.

// decimal to binary

#include<stdio.h>

#include<conio.h>

int main() {

int deci;

int r;

int sum=0;

int factor=1;

printf("enter decimal number\t");

scanf("%d",&deci);

while(deci>0) {

r=deci%2;

deci=deci/2;

sum=sum+r*factor;

factor*=10;

}

printf("The answer is %d",sum);

getch();

return 0;

}

Binary to decimal

1100=�1 ∗ 2� + �1 ∗ 2� + �0 ∗ 20+(0*2�)

// binary to decimal

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main() {



60

int bina;

int digit;

int sum=0;

int i=0;

printf("enter binary number\t");

scanf("%d",&bina);

while(bina>0) {

digit=bina%10;

bina/=10;

sum=sum+digit*pow(2,i);

i++;

}

printf("The answer is %d",sum);

getch();

return 0;

}

Exclusive use of ‘switch-case’ and ‘goto’ statements

2005PU A program that takes a sum of money and starts converting it to smaller units where

user can choose his first starting unit. The conversion is in such a way that the first starting units

takes maximum possible conversion then the left sum is supplied to conjugate term following it, this

process continues till the end.

#include<stdio.h>

#include<conio.h>

int main(){

int currency;

int choice;

int m;

printf("Enter currency\t");

scanf("%d",&currency);

printf("\n Enter 1000 to start from 1000");









printf("-----------------------------------------------\n\n\n\n");

scanf("%d",&choice);

switch (choice) {

case 1000:

goto thousand;



61

case 500:

goto fivehun;

case 100:

goto hundred;

case 50:

goto fifty;

case 20:

goto twenty;

case 10:

goto ten;

case 5:

goto five;

case 2:

goto two;

case 1:

goto one;

}

thousand:

m=currency/1000;

currency=currency%1000;

printf("The no. of thousands(1000) is\t\t\t%d\n",m);

goto fivehun;

fivehun:

m=currency/500;


printf("The no. of five hundreds(500) is\t\t\t%d\n",m);

goto hundred;

hundred:

m=currency/100;


printf("The no. of hundreds(100) is\t\t\t%d\n",m);

goto fifty;

fifty:

m=currency/50;


printf("The no. of fifty(50) is\t\t\t%d\n",m);

goto twenty;

twenty:



62

m=currency/20;


printf("The no. of twenty(20) is\t\t\t%d\n",m);

goto ten;

ten:

m=currency/10;


printf("The no. of ten(10) is\t\t\t%d\n",m);

goto five;

five:

m=currency/5;


printf("The no. of five(5) is\t\t\t%d\n",m);

goto two;

two:

m=currency/2;


printf("The no. of two(2) is\t\t\t%d\n",m);

goto one;

one:

m=currency/1;


printf("The no. of one(1) is\t\t\t%d\n",m);

getch();

return 1;

}

The general mechanism of each block of indices like ‘fifty:’ ’five hun:’….. checks whether the sum of

money left for it after preeceeding calculations; are sufficient enough to hold that value or not. If yes

then it does calculation. Here calculation means subtracting the left sum by the respective amounts



63

and then ‘goto-statement’ transfers the execution flow to next succeeding block of indices until last

index ‘one:’. Here ‘m=variable’ gives the quantity of each amount in each block. At first user is given

choice to start converting money from what bigger amount of sum like from 500 or 1000 or 100 or

50 or any. So the above mechanism starts from the respective block only. Here ‘swich-case’ is used

to meet user’s choice in various alternatives. If someone wants to coiunt money from 500 then the

sum he enters will not be counted from 1000 instead it will be counted from 500 then 100 then 50

then 20 then 10 then5 then 2 and finally 1. It is not necessary that all blocks assume some value

because the sum we enter are divisible for some; not for other and even if it is divisible the sum to

be calculated on a block changes from block to block as the calculation begins.

SORTING

Selection sorting

Algorithm (for ascending)

• Compare a [1] with second element a [2]. If a [2] is smaller than a [1] then interchange their

values.

• Compare a [1] with a [3]; interchange their values if a [3] is smaller than a [1].

• Continue comparing a [1] with other numbers.

• Repeat above processes (1, 2 and 3) for a[2],a[3]………a[n-1] comparing a[2] from a[3] to

a[n], a[3] from a[4] to a[n]………a[n-1] with a[n].

In the repeated trips the each element compares with other remaining elements coming after it and

the smallest numbers come up directly by selection (or it may step up wisely by co-incidence). Hence

it is called selection sorting as each element compares with the other. Actually, there is the

comparison of no element but there is the comparison of elements of specified addresses. Elements

don’t change, however their addresses change. So in sorting, comparison is between addresses that

take values as per condition; that is suited for the purpose-ascending or descending. Here each loop

decides the comparison of element of an address to the other addresses but not of all addresses. So

the inner loops successively give the necessary sorting. The trend of loops goes on decreasing from

(n-1) to1 because as the loop continues each time, each address gets secured on its position.

We have started comparing numbers from j=0 to j=n-1 so the address (1) gets secured and then in

next i=1 address (2) gets secured and so on. If we had started from j=n to j=2 (similarly to above

case, here j=1 automatically secured after last loop because according to the sense we know any

number of things can change their physical property; place in reference to a single thing or with

reference to a body, others can only change in regard to the values they posses, not on the specific

places they posses and hence in above case j=1 is not taken) the addresses of j=n first gets secured

then j=n-1, j=n-2……j=2 and j=1 automatically gets secured.



64

While the outer loop, is to supply each number (or actually each address) at a time which can be

altered in any positions of numbers from 1 to n-1 or n to 2. Here one is not taken this is the similar

case described above because one becomes our reference address again.

Work of outer loop (i) is to just facilitate each numbers each time except the reference address.

Work of inner loop (j) is to compare each address (pointer) with the other taking a reference

address. The result won’t alter if we take under consideration- the comparison of reference address

but a good program minimizes the working time and process and in other proper understanding of

pointers is the input for this program. So we don’t take the reference pointer under consideration.


Illustration of Selection Sort of four items

In each trip, a value gets secured (or a pointer gets secured). The number of comparisons= (n-1) + (n-

2) + (n-3) +………1 Sum of (n-1)……….1 is n (n-1)/2

[NOTE: Actually, we directly don’t declare reference pointer but on our assumptions, automatically

the reference pointer is achieved. We need to analyze it for more efficiency of the program.]



65

For descending we can simply change the condition; inverse the condition.

NOTE: Examples are done in coming using selection and bubble sorting.

Bubble sorting

For ascending,

• Compare first two elements a [1] and a [2]. If a [2] is smaller than a [1], interchange their

values.

• Compare a [2] and a [3]; interchange them if a [3] is smaller than a [2].

• Continue for all values serially.

• Repeat above (1, 2, and 3) for (n-1) times.

In repeated trips through the array, the smallest elements ‘bubble up’ to the top one step at one

loop (inner loop). Because of this bubbling up effect, this algorithm is called bubble sorting.

Ilustratration of Bubble Sort of four items

In each trip, the smallest number ‘10’ shifts one position up at the end of trips. The number of

comparisons = n (n-1)/2

In above,



66

4(4-1)/2 = 3+2+1

6=6 proved

For Descending, change

if (a[j]>a [j+1]) { to if (a[j] <a [j+1]) {

Third sorting process

The number of loops for both bubble and selection remains the same. The sorting processes are the

same for both string and numbers. The above two sorting processes worked on the basis that each

number is compared with others in the order of (n-1) to (n-2) to (n-3) to (n-4)………….to 1 times. This

may sometime give unnecessary processes that take long time consuming in the case of dealing with

a large quantity of numbers. Co-incidentally two or more numbers sometime may be in ascending or

descending before our interaction with them. This ascending or descending may be with numbers of

small packet range to higher packet range (here packet means a group i.e. numbers in consequent

places) or these packets may be in large quantities i.e. more in numbers quantitatively. However the

above programs cannot take the consideration or advantage of this (if it is available by co-incidence)

because the above program works taking the base of pointer first then the numbers in pointers.

There two loops were used. This advantage can be used by a concept using goto and a single loop.

The above programs worked on the basis of securing each pointer each time and decreasing the

inner loops, each time of outer loop. Taking both the advantages one of decreasing inner loop and

other of taking probability of packets of ascending or descending numbers( that may have high

probability of low or 0 co-incidentally), we can solve this by taking our base with the numbers (i.e.

their values) at first then the pointers that are connected to. This is as follow:

for(i=0;i<n;i++) {


A:

for(j=0;j<i;j++) {

if(num[j]<number) {

temp=num[j];

num[j]=number;

number=temp;

j++; /*deleting these two statements ‘j++’ ,‘goto A:’ also doesn’t affect the output of program*/

goto A;

}

}

}

for(i=0;i<5;i++) {

printf("%d",num[i]);

}

This method is exclusively used in coming sorting program.

Analysis of the solution

1. Numbers fixed

2. Numbers dynamic: - Dynamics over come and its special case is static. So, we take

the dynamic array analysis. Each time a number is taken from users. Before that, array with

proper dimensions and sizes are declared. So there are vacant spaces for the numbers. It



67

doesn’t matter whether the consideration with string, character or numbers. Let us consider

the above program for 1-D and for numbers. First of all the loop under (i) takes each number

from user but it is not assigned to array because if it is assigned a next loop will be required a

more time and this is with basis of pointer rather than number similar to above sorting

processes and hence it doesn’t work on our requirement. Assigning to an array means basis

on pointers rather than numbers. So the number is stored in variable called number that

takes different values supplied for different times or in different loops of (i). Then the second

loop of (j) starts here ‘A:’ has important to do (to be discussed in coming parts). The number

is compared to num[j] (initially 0) and hence the number is assigned to num[j] (i.e. initially

num [0]). This is a special case that works for the first number supplied by user. Now num[j]

is not vacant.

‘Interchanging values’ is done in above sections, refer it for analysis.

Here number supplied gets its position (but not secured positions because there are still many

numbers to be supplied by users and to be compared by program). At i=0, j=0, number takes value of

temp; value of original num[j] =0. So it goes to ‘A:’ label but condition fails( and it is intended to fail)

because num[j]<numbervalue supplied initially <0 (which failed for positive value), if the value is

negative the case will be other because if the number is negative then during first condition before

‘goto’ the condition fails the negative number supplied goes to vain. There is no space, no array for

negative numbers because the num[j] =0 automatically (initially). So our program works with the

assumption that the first number should be positive but the second, third……numbers can be both

negatives and positives. However we can take advantages of this case for the case of positive

numbers. We can remove the disadvantage by taking a very negative number as num [0] and

executing all but in this case the output of sorted numbers should be from num [0] to num [n-1] as

regular in above. This is for the case of descending.

Now the number zero (declared automatically) goes to vain. This was intended to do. J=0 to I j=0to0

and the inner loop exits and i=1 so that a number is taken from user compared to num[0] if number

is big it replaces it and values interchange and ‘goto’ makes to take the other value to other array

element. In this way, if the condition of ‘if’ is satisfied then it displace the value of certain array index

and the value of array index is tested for array [index+1] so that the values get proper place for their

descending values.



68

Here the value of j is incremented for j, because the displaced numbers need to be compared with

the index after it which is sure to be displaced by it because the process is in descending at last of

each loop. In this way, our achievement is got.

Advantage → Sorting is based on each number so the sorting is purely dynamic, each case and each

end of loop gives required sorting of numbers. This sorting even in transition period can be useful for

our various purposes and projects.

Disadvantage → If sorted packets aren’t available in sufficient way or it can’t overcome the pointer

basis sorting (that may be due to less probability in gradual sequential deviation) then the number of

loops is high and time consuming.

In all cases descending and ascending are changed by inversing the condition within if (). But in third

sorting for descending, num [0] is declared zero it can also be left doing nothing as with ascending

(depending upon compiler).

PU2011FALL 5b)

Sort ‘n’ numbers by function (special)

/*

we are sorting in ascending order,

this process is a type of third sorting we described in sorting-section



69

*/

#include<stdio.h>

#include<conio.h>

int i,j,number,num,temp;

int innersort[100]; //declare it globally

void sorting(int num);

void printing();

void main(){

printf("how many numbers?");


for(i=0;i<number;i++){

printf("\nEnter a number:\t");

scanf("%d",&num);

sorting(num);// call the function for each value entered

}

printing();

getch();

}

void sorting(int num) {

static int counter=0; //can be declared as global variable too

printf("num to be ckd=%d\n\n",num);

for(j=0;j<counter;j++) {

if(num<innersort[j]) {

temp=innersort[j];

printf("one swapping done\t"); /*this line shows only execution (swap occurs) if entered new

number is less than previously entered numbers and the number of execution of this statement for a

newly entered number shows that the number of times the swapping is done for the new number

with the stored numbers*/

innersort[j]=num;

num=temp;

}

}

innersort[counter]=num;

counter++;

}

void printing(void) {

printf("\nsorted list is\n");

for(i=0;i<number;i++){

printf("\n%d",innersort[i]);

}

}



70

Description

It is good if you look the 3@=sorting type described in the sorting portion. This above technique takes

each number from user and does ‘innersort’-function invoke. This function takes each number as

input and sort in the series of previously sorted list. To count the previously saved sorted numbers

counter variable (static type is used here so that its value has scope within function and the

declaration static counter=0 is used only in 1KL time function calling, later in coming calls the

statement counter++ affects it only increasing by one per time the new number is given as argument

to function but … it doesn’t hamper if we use counter a global variable in main method. Similar as

counter we can declare innersort[100] as static type inside function but the problem is that at last

while printing the sorted output we need to print the values of innersort which is done by printing()

function as it can’t use static innersort() declared inside sorting function, so extern or globally the

variable innersort is declared) is used. The loop of I starts from 0 to counter-1. The input num is

checked to array of ‘innersort[]’, if condition is met then the num replaces the particular innersort[]’s

value and the innersort[]’s value is taken by num and the loop still continues till all numbers are

checked in condition. Here the new ‘num’ checks the values for the left part of i-loop and at last,

after successive replacing appropriate values at appropriate places there is left a value num whose

address of address of array, so the ‘innersort[counter]=num’ gives the last position to the number.

Counter is increased each time as each new ‘num’ is assigned a new place so each time a number is

increasing. The declaration of counter, increment, scope using of counter in specific positions of the

program has its own vital and specific purposes. Initially counter=0 so that the loop of i fails and it is

our intention too because at initializing 1KLnumber there is no need of any sorts of comparisons.

Then the number sis initialized to array index0; the 1KLposition then counter is incremented showing

that how many numbers are sorted in the list. Hence the value of counter at various points shows

the number of elements sorted and added in the list of innersort array.

But after a condition is met, the number from innersort and assigned to num has nothing to do with

the left upcoming pattern of sorting in i-loop except finding a place between number replacing its

origin place and the number previously succeeding it. So we can use a method below:-

For e.g.

1 20 50 100



71

If is the list of innersort[] elements.

If the new ‘num’ is then it must be inserted in between and as like

Index 0 1 2 3 resulting into

Index 0 1 2 3 4

Such direct replacement is not possible in C, so we continue our while loop of i and replace in each

loop and at last a new position is declared.

Let us analyze the problem in reality

Solution:

Here 4 numbers 1, 20, 50, and 100 are already declared in inner loop as

innerloop[0]=1

innerloop[1]=20

innerloop[2]=50

innerloop[3]=100

Now the function ‘sorting()’ is called; let us suppose for a new value 25 as sorting(25).

Here the value of counter =4 because as we know counter=number of values sorted in innersort[] as

described earlier. So the remaining process goes like for (i=0 to 3)

i Condition Result Remark

0 25<1 False Just continue loop of i

1 25<20 False Just continue loop of i

2 25<50 True Replace num and innersort[2]’s value i.e.

temp=20;

innersort[2]=25;

1 20 50 100

25

1

20 20 50

20 25 50 100



72

num=50;

(cont…..)

So the list now in innersort looks like

Index 0 1 2 3

Value

and the variable num=50.

Now this variable needs to be inserted but it doesn’t matter us because there is still a loop of i to be

continued.

i Condition Result Remark

3 50<100 True Replace num and innersort[3]’s value i.e.

temp=100;

innersort[3]=502;

num=100;

Now the list of innersort looks

Index 0 1 2 3

Value

and num=100.

Now it is directly assigned by innersort[counter] i.e.

innersort[4]=100;

and the list is as

Index 0 1 2 3 4

Value

This is our required aim of function. In this way the function maintains the sorting ability for each

new number taken each time from user.

A SPECIAL ANALYSIS

We saw that after we have replaced 25 in its position the new num for the particular instant was 50

and for next loop was 1 needs no further comparison in between themselves i.e. after the

comparison of num with the sorted list and getting its appropriate the other numbers need o other

comparisons as they were sorted in previous sorting i.e. of previous function call. So, we can just

change their address position’s index by +1, like below:-

1

1

1

20 25 100

20 25 50

20 25 50 100



73

void sorting(num){

for(i=0;i<counter;i++) { //i=0 to counter-1

if(num<innersort[i]) {

temp[counter-1-i];

temp[0]=innersort[i];

innersort[i]=num;

for(j=1;j<counter-i;j++) { //j=1 to counter-1-i

temp[j]=innersort[i+j];

}

for(k=0;k<counter-i;k++) { //k=0 to counter-1-i

innersort[i+k+1]=temp[k];

}

printing();//break the function

}

}

The declaration

temp[counter-1-i]; why?

counter show total positions but as the array starts from zero index so -1 is used as i times the loop

is runned already or the particular instant value of i resembles the number of index of innersort up

to which value is sorted.

Then this ‘temp-array’ assigns all the values including the value replaced by the num to all the

succeeding values.

We take advantage of value of j in finding the left positions of values and assigning it to temp

governed by j. so j has two purposes, one finding position of preceding values and assigning to new

temp values.

At last k-loop is run,

innersort[i+k+1]=temp[k] ?

The value of i is of index and similarly is with k, but the position value differs by +1 as we need the

next position to the index i so at first i+k+1=i+0+1=i+1, next time i+1+1=i+2 and so on, the value is

index resenting temp[]. As value of k increases, the new position of innersort is assigned a new value

at each time of increment.

Now we are adding a new feature to above program

/*

we are sorting in ascernding order,

this process is a type of third sorting we described in sorting-section

*/

#include<stdio.h>

#include<conio.h>

int i,j,num,temp,counting_num=0;

int innersort[100];

void sorting(int num);

void printing(void);

void main() {



74

readnextnumber:

printf("Enter any number\nOr enter -999 to look sorting:\t");

scanf("%d",&num);

if(num==-999){ goto outside;}

counting_num++;

sorting(num);

goto readnextnumber;

outside:

printing();

getch();

}

void sorting(int num) {

static int counter=0;

for(j=0;j<counter;j++) {

if(num<innersort[j]) {

temp=innersort[j];

innersort[j]=num;

num=temp;

}

}

innersort[counter]=num;

counter++;

}

void printing(void) {

printf("The sorted list is\n");

for(i=0;i<counting_num;i++){

printf("%d\n",innersort[i]);

}

}

Here we are not directly asking how many number to be sorted instead we are saying that -999

made as an input shall determine the end of input automatically.

EXAMPLES OF SORTING

SORT NUMBERS ASCENDING (SELECTION SORTING)



75

//SORT NUMBERS ASCENDING (SELECTION SORTING):

#include<stdio.h>

#include<conio.h>

int main() {

int i=0,j=0,number[100],temp,n;

printf("Enter how many numbers\t");

scanf("%d",&n);

printf("\n\n\n");

for(i=0;i<n;i++) {

printf("Enter [%d] number\t",i+1);


}

/*Sorting begins-ascending*/

for(i=1;i<n;i++) {

for(j=i+1;j<=n;j++) {

if(number[i]>number[j]) {

temp=number[i];

number[i]=number[j];

number[j]=temp;

}

}

}

printf("\n Ascending List \n\n");

for(i=0;i<n;i++)

printf("%d\t",number[i]);

getch();

return 0;

}

SORT NUMBERS DESCENDING (BUBBLE SORTING)

//SORT NUMBERS DESCENDING (BUBBLE SORTING)

#include<stdio.h>

#include<conio.h>

int main() {

int i=0,j=0,number[100],temp=0,n;


scanf("%d",&n);



76

printf("\n\n\n");

for(i=0;i<n;i++) {



}


for(i=0;i<n;i++) {

for(j=(i+1);j<n;j++) { //for(j=1;j<=n-i;j++)can be used as at first point j=(i+1)=(0+1)=1 and j<=n-i is

same to j<n

if(number[i]<number[j]) {

temp=number[i];


number[j]=temp;

}

}

}

printf("\n Descending order \n\n");

for(i=0;i<n;i++)

printf("\t%d\t",number[i]);

getch();

return 0;

}

SORT 5 NAMES ASCENDING (BUBBLE SORTING)

Here sort is necessary we use bubble sort or selection sort or the last sort we looked in sort portion,

similarly here dummy; a temporary variable works as temp variable of before whose only purpose is

to interchange values. Strcpy() copies string [j-1] (string[0], string[1], string[2]…….)to dummy which is

taken by string[j] (string[1],string[2]……) All its process are similar to the above three cases of

numeric.

//SORT NAMES ASCENDING (BUBBLE SORTING)

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

char string[100][20],dummy[20];

int i=0,j=0,n;

printf("Enter quantity\n");

scanf("%d",&n);

while (i<n) {

printf("Enter a name\t:");

scanf("%s",&string[i++]);

} //while-loop terminal


for(i=1;i<n;i++) {

for(j=1;j<=n-i;j++) {

if(strcmp(string[j-1],string[j])>0) {

strcpy(dummy,string[j-1]);



77

strcpy(string[j-1],string[j]);

strcpy(string[j],dummy);

}

}

}

printf("\n Alphabetical List \n\n");

for(i=0;i<n;i++)

printf("\n[%d]=\t%s",i+1,string[i]);

getch();

}

SORT NAMES DESCENDING (SELECTION SORTING)

//SORT NAMES DESCENDING (SELECTION SORTING)

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

char string[100][20], dummy[20];

int i=0,j=0;

int num;

printf("Enter qunatity:\t");

scanf("%d",&num);

printf("Enter names of %d items\n",num);

while (i<num) {

printf("Enter a name");

scanf("%s",&string[i++]);

}

for(i=0;i<(num-1);i++) { //the last is rejected automatically

for(j=i+1;j<num;j++) {

if((strcmp(string[i],string[j]))<0) {

strcpy(dummy,string[j]);

strcpy(string[j],string[i]);

strcpy(string[i],dummy);

}

}

}

printf("\n Descending List \n\n");



78

for(i=0;i<num;i++)

printf("\n[string%d]=%s",i+1,string[i]);

getch();

}

A PROGRAM THAT TAKES NAMES AND FIVE SUBJECT MARKS OF FIVE STUDENTS THAT

DISPLAY STUDENT RANKS AND SUBJECT MARKS (BY USING SELECTION SORTING)

Hints:

Here array is an important too to store names, their corresponding subject marks, and its

percentages. We sort percentage, in the way of sorting percentage, we interchange corresponding

values of names so that in the way of our sorting (in this case descending-for mark sheet) percentage

it corresponds to its names and hence the two field names and percentage interchange

simultaneously. If we want to print subject marks, then we need to interchange subject mark’s

pointer similar to the above cases.

Median and Standard Deviation

For median

We first arrange in ascending or descending order then

For even quantity of numbers, median=�G/�NO7�P�70NO

� value

For odd, median=�P�70NO

� value

//median

#include<stdio.h>

#include<conio.h>

int main() {

int i=0,j=0,number[100],temp;

int n;

int median;


scanf("%d",&n);

printf("\n\n\n");

for(i=0;i<n;i++) {



}


for(i=1;i<n;i++) {

for(j=i+1;j<=n;j++) {

if(number[i]>number[j]) {



79

temp=number[i];


number[j]=temp;

}

}

}

if(n%2==0) {

median=(float)(number[n/2-1]+number[n/2+1-1])/2;

/*here in contrast to formula,'-1', 1 is subtracted at last ascharacter array starts from zero so

to maintain equilibrium with such array*/

}

else {

median=number[n/2+1-1]; /*same to above*/

}

printf("the median is %d",median);

getch();

return 0;

}

For standard deviation

We find mean; Mean=0G ∑ xG0

then variance is find out

Variance= 0G ∑ �x − mean�G0

Standard deviation=√variance

//mean, variance and standard deviation

#include<stdio.h>

#include<conio.h>

#include<math.h>

int main() {

int i=0,number[100];

int n;

float mean=0,variance,sd;


scanf("%d",&n);

printf("\n\n\n");



80

for(i=0;i<n;i++) { /*input and mean calculation*/



mean=mean+(float)number[i]/n;

/*as we know total quantity then mean is the sumation of all numbers divided by total

quantity*/

/*here casting is essential to prevent truncation of value (losing values afetr deciaml)due to

integer division*/

}

printf("The mean is \t%.4f\n",mean);

for(i=0;i<n;i++) { /*variance calculation*/

variance=(float)1/n*(pow(((float)number[i]-mean),2));

/* instead we can do divide by 'n' after all summation afetr for-loop too*/

/*casting is essential*/

}

printf("The variance is \t%.4f\n",variance);

sd=pow(variance,0.5); /*standard deviation calculation*/

printf("The standard deviation is \t%.4f",sd);

getch();

return 0;

}

MATRIX ARRAY

Matrix is 2-D like in similar way the array elements are actually stored (whether it is 2-D or 3-D or

multi-D). So taking this advantage matrix can be easily expressed in terms of array elements where

loops are used to define and express array elements (detailed in Array Section). Here define means

declaration and intake whereas express means printing or analyzing array elements for calculations

or comparisons.



81

CASES:

Case-1: Finding smallest and greatest of a matrix

For this we keep the first value to ‘small’ and ‘big’ and compare each value according to required

and if condition satisfies the new value is assigned to the variables and lastly we print output.

//greatest and smallest element in matrix

#include<stdio.h>

#include<conio.h>

void main(){

int matrix[3][3];

int small,big;

int i,j;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

printf("Enter matrix1[%d][%d]",i,j);

scanf("%d",&matrix[i][j]);

}

}

small=matrix[0][0];

big=matrix[0][0];

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

if(matrix[i][j]>big) big= matrix[i][j];

if(matrix[i][j]<small) small= matrix[i][j];

}

}

printf("The smallest matrix element is %d\n",small);

printf("The biggest matrix element is %d",big);

getch();}

Case-2:

Sum of diagonal elements of square 3*3 matrix

1. sum of diagonal from right top to left bottom

1 2 34 5 6 7 8 9 here required = 3+5+7

3=matrix[0][2]

5= matrix[1][1]

7= matrix[2][0] i.e the value of sum of indices is less than 3 or equal to 2



82

//sum of diagonal from right top to left bottom

#include<stdio.h>

#include<conio.h>

void main(){

int matrix[3][3];

int i,j;

int sum=0;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {



}

}

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

if(i+j==2){ //i+j<3 also works

sum=sum+matrix[i][j];

}

}

}

printf("The sum of diagonal from right top to left bottom %d\n",sum);

getch();

}

2. sum of diagonal from left top to right bottom

1 2 34 5 6 7 8 9 here required = 1+5+9

3=matrix[0][0]

5= matrix[1][1]

7= matrix[2][2] i.e the value of index ‘I’= index’j’

//sum of diagonal from left top to right bottom

#include<stdio.h>

#include<conio.h>

void main(){

int matrix[3][3];

int i,j;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {



}

}

int sum=0;



83

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

if(i==j){


}

}

}

printf("The sum of diagonal from left top to right bottom %d\n",sum);

getch();}

Case-3: Finding a matrix element in a matrix

For this, comparison is done with all values or elements of matrix and lastly result is shown. If it is

found then counter =1 and hence we don’t print “not found”.

// Finding a matrix element in a matrix

#include<stdio.h>

#include<conio.h>

void main(){

int matrix[3][3];

int i,j;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {



}

}

int search;

printf("Enterthe element to be find out?\t");

scanf("%d",&search);

int counter=0;

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

if(matrix[i][j]==search){

printf("the element [%d] is found in [%d] row and [%d] column",search,i+1,j+1);

counter+=1;

}

}

}

if(counter!=1){

printf("not found");

}

getch();}

Case-4: sum of all elements in a matrix



84

Here’ sum’ sums all the values of matrix.

//Case-3: sum of allelements in a matrix

#include<stdio.h>

#include<conio.h>

void main(){

int matrix[3][3];

int i,j;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {



}

}

int sum=0;

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {


}

}

printf("the sum is %d",sum);

getch();

}

Case-5: transpose of a matrix

Here we know in transpose matix the values are interchanged in rows and columns that is I and j are

interchanged.

1 2 34 5 6 7 8 9

Its transpose is

1 4 72 5 8 3 6 9

// transpose of matrix

#include<stdio.h>

#include<conio.h>

int main(){

int matrix[3][3],newmatrix[3][3];

int i,j;

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {



}

}



85

int sum=0;

printf("\n\n");

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

newmatrix[j][i]=matrix[i][j];

}

}

for(i=0;i<3;i++) {

for(j=0;j<3;j++) {

printf("%d\t",newmatrix[i][j]);

}

printf("\n");

}

getch();

}

Matrix Multiplication

of n×m and m×p orders →n×p orders matrix

Here the new matrix is of order n ×p. So the outer loop is from i=0 to n-1, second loop is similarly

i=0 to p-1, then an inner loop is for looping till the required sum is obtained. This sum is obtained

from summing column elements of second matrix i.e. p times to the row elements of first matrix. It

doesn’t matter how many row elements are considered because we have earlier declared the

positions of new matrix by first and second loops. Third loop is just for sum. So third loop is from k=0

to m-1. The obtained final matrix is of order n×p.

matrix1[n][m]

matrix2[m][p]

matrix3[n][p]

for i= 0 to n-1 {

for j= 0 to p-1 {

sum=0

for k = 0 to m-1 {

sum=sum+matrix1[i][k]*matrix2[k][j];

}

matrix3[i][j]=sum

}

}

Here i=0 to n-1 (and similarly for j, k) because in array addresses starts from 0 to n-1 for n elements.

By matrix multiplication, as element of new matrix has the row equal to first matrix and column

equal to second matrix and during multiplication there is the sum of elements in such that first

matrix row remains same and column changes in such a way that it equals to second row and its

column constant. This is illustrated above. There first matrix’s row i=constant, second matrix column



86

j= constant but first’s column and second’s row changes k=not constant even within loop of j. This is

basic principle of matrix multiplication of any order matrix to any order matrix. This can be simplified

to 3×3 matrix × 3×3 matrix, for this m=n=3.

If inputs are

1 2 34 5 6 7 8 9 and

1 1 11 0 00 0 1 then output is 3 1 49 4 1015 7 16

//MATRIX MULTIPLICATION OF (3*3) AND (3*3) MATRICES

#include<stdio.h>

#include<conio.h>

#define N 3

int main(){

int matrixa[3][3],matrixb[3][3],matrixc[3][3];

int sum=0,i,j,k;

printf("MATRIX MULTIPLICATION OF (3*3) AND (3*3) MATRICES\n\n\n");

printf("\n\n");

for(i=0;i<N;i++) {

for(j=0;j<N;j++) {

printf("Enter matrixa[%d][%d]",i,j);

scanf("%d",&matrixa[i][j]);

}

}

printf("\n\n");

for(i=0;i<N;i++) {

for(j=0;j<N;j++) {

printf("Enter matrixb[%d][%d]",i,j);

scanf("%d",&matrixb[i][j]);

}

}

for(i=0;i<N;i++){

for(j=0;j<N;j++){

sum=0;

for(k=0;k<N;k++){

sum=sum+((matrixa[i][k])*(matrixb[k][j]));

}

matrixc[i][j]=sum;

}

}

printf("\nNew matrix3\n");

for(i=0;i<N;i++) {

for(j=0;j<N;j++) {

printf("%d\t",matrixc[i][j]);

}

printf("\n");

}

getch();

return 1;

}



87

More generalization of matrices multiplication

//MATRIX MULTIPLICATION OF (n*m) AND (m*p) MATRICES

#include<stdio.h>

#include<conio.h>

void main(){

int matrix1[100][100],matrix2[100][100],matrix3[100][100],n,m,p,sum,i,j,k;

printf("MATRIX MULTIPLICATION OF (n*m) AND (m*p) MATRICES\n\n\n");

printf("Enter value of n");

scanf("%d",&n);

printf("Enter value of m");

scanf("%d",&m);

printf("Enter value of p");

scanf("%d",&p);

printf("\n\n");

for(i=0;i<n;i++) {

for(j=0;j<m;j++) {


scanf("%d",&matrix1[i][j]);

}

}

printf("\n\n");

for(i=0;i<m;i++) {

for(j=0;j<p;j++) {


scanf("%d",&matrix2[i][j]);

}

}

for(i=0;i<n;i++){



88

for(j=0;j<p;j++){

sum=0;

for(k=0;k<m;k++){

sum=sum+matrix1[i][k]*matrix2[k][j];

}

matrix3[i][j]=sum;

}

}

printf("\nNew matrix3\n");

for(i=0;i<n;i++) {

for(j=0;j<p;j++) {

printf("%d\t",matrix3[i][j]);

}

printf("\n");

}

getch();

}

TASK: A simple program that takes students marks in various subjects, name and roll numbers to

print out ranks, roll numbers and marks (both % and marks) in Descending order.

Hints: Use concept of array in such a way that the marks, percentages, names and roll numbers

while in sorting by descending by any method to marks-percentage corresponds to same change in

every other variable so that name, marks, percentage, roll and rank corresponds to required

particular person.

You may add search by roll or search by few initial words options. For this use string function to

compare the input to available name values stored and in true then print output. For few initial

words firstly find string-length so that comparison is done to only that much length of names stored.

Solution:

//a program taking student's marks and printing out rank without using file handling and just to

array for beginners

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

int count=0;

int roll[100];

char name[100][100];

float mark1[100],mark2[100],percent[100];

int main() {

int num;

int number,a;

int temp_count=count;

char dummy[20];

int check;

char checkk[30];



89

int j=0,k=0; float temp=0; int temp_i=0;

A:

printf("What do you want to do?\n");

printf("[1]\tto add mark list\n");

printf("[2]\tto look mark list\n");

printf("[3]\tto search by roll\n");

printf("[4]\tto search by name\n");

printf("[5]\tto exit\n");

printf("-------------------------------------------\n");

scanf("%d",&num);

printf("-------------------------------------------\n");

switch(num) {

case 1:

printf("Enter the number of input student's marks?\t");


for(a=temp_count;a<(number+temp_count);a++) {

printf("Enter Roll\n");

printf("Enter Name(ENTER Single name)\n");

printf("Enter Marks in Computer\n");

printf("Enter Marks in Maths\n");

scanf("%d",&roll[a]);

scanf("%s",&name[a]);

scanf("%f",&mark1[a]);

scanf("%f",&mark2[a]);

percent[a]=(mark1[a]+mark2[a])/2;

count++;

}

if(getch()) goto A;

case 2:

printf("MARKSHEET OF STUDENTS\t(RANK WISE)\n\n");

printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");

printf("Computer ");

printf("Maths ");

printf("PERCENT");

//Sorting begins-descending by bubble method

for(j=1;j<count;j++) {

for(k=1;k<=count-j;k++) {

if(percent[k-1]<percent[k]) {

temp_i=roll[k-1];

roll[k-1]=roll[k];

roll[k]=temp_i;

strcpy(dummy,name[k-1]);

strcpy(name[k-1],name[k]);

strcpy(name[k],dummy);



90

temp=mark1[k-1];

mark1[k-1]=mark1[k];

mark1[k]=temp;

temp=mark2[k-1];

mark2[k-1]=mark2[k];

mark2[k]=temp;

temp=percent[k-1];

percent[k-1]=percent[k];

percent[k]=temp;

}

}

}

// Rank List

int i;

for(i=0;i<count;i++) {

printf("%d\t",i+1);

printf("\n%d\t",roll[i]);

printf("%s\t",name[i]);

printf("%f\t",mark1[i]);


printf("%f\n",percent[i]);

}

if(getch()) goto A;

case 3:

printf("Enter roll number\t");

scanf("%d",&check);

for(i=0;i<count;a++) {

if(roll[i]==check){


printf("\n\n----------------------------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");

printf("PERCENT\n");

printf("%d\t",i+1);

printf("%d\t",roll[i]);




printf("%f\n\n",percent[i]);

break;

}

}

if(getch()) goto A;

case 4:

printf("Enter name\t");



91

scanf("%s",&checkk);

for(i=0;i<count;a++) {

if(strncmp(name[i],checkk,strlen(checkk))==0){

printf("hello");


printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");

printf("PERCENT\n");

printf("%d\t",i+1);

printf("%d\t",roll[i]);




printf("%f\n\n\n",percent[i]);

break;

}

}

if(getch()) goto A;

case 5:

printf("THANKS TO YOU AND HOPE YOU TO LOOK SIMILAR USING FIHE HANDLING IN COMING

LESSONS");

if(getch()){

exit(0);

}

}

getch();

return 1;

}

RECURSIVE FUNCTION

It is a method for dividing a problem into smaller units in such a way that each units are similar to

other irrespective of some small changes that can be made change gradually and eventually to al so

that a function is called for required number of times within another function, the last condition of

function to be called or not must be declared for e.g. in factorial



92

if(n==1) return (1);

is last condition that terminates the function. A function is repeatedly called to individual thje

problem into parts however a function is called within another function so there is the combination

of results to give a net output.

To calculate number of permutations and combinations using function factorial

For permutation

n[\n!�n − r!

For combination

n]\n!�n − r! r!

#include<stdio.h>

#include<conio.h>

int factorial(int n) {

int fact=1; int i;

for(i=1;i<=n;i++)

fact=fact*i;

return fact;

}

int main() {

int n,r;

printf("n\t");

scanf("%d",&n);

printf("r\t");

scanf("%d",&r);

int f1,f2,f3;

f1=factorial(n);

f2=factorial(n-r);

f3=factorial(r);

int com=f1/(f2*f3);

int per=f1/f2;

printf("The combination is %d",com);

printf("\n The permutation is %d",per);

getch();

return 0;}

Factorial by recursive

#include<stdio.h>

#include<conio.h>



93

int factorial(int);

void main() {

int num;


scanf("%d",&num);

printf("The factorial of %d is %d\t",num,factorial(num));

getch();

}

int factorial(int num) {

if(num==1){

return(1);

}

else

{

return(num*factorial(num-1));

}

}


1 1 2 3 5 8 13 21 ……………….

#include<stdio.h>

#include<conio.h>

int fibo(int);

int main() {

int num;


scanf("%d",&num);

printf("The fibonaci \t");

int i;

for(i=0;i<num;i++) {

printf("\t%d",fibo(i+1));

}

getch();

return 1;

}

int fibo(int num) {

if(num==1){

return(1);

}

else if(num==2){

return(1);

}

else

{

return(fibo(num-2)+fibo(num-1));

}

}



94


0 1 1 2 3 5 8 13 21 ……………….

#include<stdio.h>

#include<conio.h>

int fibo(int);

int i;

int main() {

int num;


scanf("%d",&num);

printf("The fibonaci \t");

for(i=0;i<num;i++) {

printf("\t%d",fibo(i+1));

}

getch();

return 1;

}

int fibo(int num) {

if(num==1){

return(0);

}

else if(num==2){

return(1);

}

else

{

return(fibo(num-2)+fibo(num-1));

}

}

To calculate power by recursive

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<stdlib.h>

float power(float x,int n){

if(exp<0) {exit(1);}

else if (n==0) { return (1);}

else return (x*power(x,n-1));

getch();

}

void main() {

float x;

int n;


scanf("%f",&x);

printf("enter power \t");

scanf("%i",&n);



95

printf("The answer is %f",power(x,n));

getch();

}

Calculate sum of digits by recursive

#include<stdio.h>

#include<conio.h>

extern suma=0;

int sum(int num){

int i=1,digit=0;

digit=num%10;

suma+=digit;

num/=10;

i++;

if(num>0){

sum(num);

}

return (suma);

}

void main() {

int number;



printf("The answer is %d",sum(number));

getch();

}

Armstrong using function

#include<stdio.h>

#include<conio.h>

#include<math.h>

int arms(int num) {

int sum=0;

int digit=0;

do {

digit=num%10;

num=num/10;sum=sum+pow(digit,3);}

while (num>0);

return sum;

}

int main() {

int num;


scanf("%d",&num);



96

if(arms(num)==num) {

printf("yes it is armstrong");

}

else printf("not");

getch();

return 1;

}

Armstrong using recursive

#include<stdio.h>

#include<conio.h>

#include<math.h>

int arms(int num) {

int sum=0;//must be inside function arms() so that for each recursive it starts from 0 so it is local

int digit;

printf("the num is %d\n",num);

printf("the sum is %d\n",sum);

if(num==0){

return 0;}

else {

digit=num%10;

num=num/10;

sum=pow(digit,3);

return(sum+arms(num));

}

}

int main() {

int digit;

int total=0;

int a;

int num;


scanf("%d",&num);

if(arms(num)==num) {

printf("yes it is armstrong");

}

else printf("not armstrong");

getch();

return 1;

}

WITHOUT FUNCTION

To check token without using any function



97

#include<stdio.h>

#include<conio.h>

int main() {

printf("enter a token:\t");

char b;

scanf("%c",&b); //address '&' is compulsory here

int a=b;

if(a>=48 && a<=57) {

printf("digit");

}

else if(a>=65 && a<=90) {

printf("Ucase");

}

else if(a>=97 && a<=122) {

printf("Lcase");

}

else {

printf("special");

}

getch();

return 0;

}

Calculate power without function

//calculate power

#include<stdio.h>

#include<conio.h>

int main() {

int base;

printf("enter base\t");

scanf("%d",&base);

int power;

printf("enter power\t");

scanf("%d",&power);

int i;

int result=1;;

for(i=1;i<=power;i++){

result=result*base;

}

printf("The answer is %d",result);

getch();

return 0;

}

To reverse a sentence without any function



98

Here length of string is calculated by while loop that increments varable length staring from zero

until we read ‘\0’ i.e. the termination of string, while reading each character of string by help of

counter ‘i’ that increase for each loop of while, the value of i starting from zero that measn reading

character of string from character array position zero.

Then i= length-1 cause the no of index in character array ends to length-1. j loop starts from 0 to

length-1 where I continuously increases that is reading position of string increases in the other hand

the character to be signed is done by j in such a way that it fills the last space of rev string

continuing to fill up to first.

Here j=from 0 to length-1 and i=length-1 to 0

So the loop exactly matches to both reading and writing material.

// To reverse a sentence without any function

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

int j;

char rev[100];

int length=0,i=0;

char input[100];

printf("enter a sentence\t");

gets(input);

while (input[i]!='\0')

{

length++; i++;

}

i=length-1;

for(j=0;j<length;j++,i--)

rev[j]=input[i];

rev[j]='\0';

printf("%s",rev);

getch();

return 0;

}

Concatenate two strings without function

// this process is exactly similar to working-mecchanism of strcat() function

Here similar to above process we calculate the length of two strings s1 and s2. The value of i=0 after

the length of s1 is found so that no new variable is needed.



99

After that a new variable cona takes all values of s1 one by one by character array method where

i=0to length of l1-1 then in second loop i=length of l1 to less than sum of both length because the

adding process in cona must start from the ending of last loop of i so and this second loop continues

for length of l2 times if count from 1. So the loop starts from l1 to less than (l1+l2). Finally ‘\0’ is

added to make a string.

//Concatenate two strings without function

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

char s1[100],s2[100];

printf("enter a string\t");

gets(s1);

printf("enter another string\t");

gets(s2);

int l1=0,l2=0,i=0;

while (s1[i]!='\0')

{

l1++; i++;

}

i=0;

while (s2[i]!='\0')

{

l2++; i++;

}

char cona[200];

for(i=0;i<l1;i++){

cona[i]=s1[i];

}

int j=0;

for(i=l1;i<(l1+l2);i++,j++)

cona[i]=s2[j];

cona[i]='\0';

printf("%s",cona);

getch();

return 0;

}

Search specific word in given string

Here

for(i=0;i<=(l1-l2);i++){ // the maximum possible condition to find word

this loop starts from zerp to maximum possible to find the word



100

for e.g if ‘ef’ is to be find in ‘abcdef’ then five loops must go on each loop’s length equal to length

‘ef’ that is 2

.loops from 0 to 4

For loop0 compare ef to ab

For loop1 compare ef to bc

For loop2 compare ef to cd

For loop3 compare ef to de

For loop4 compare ef to ef

The fifth loops is thus the maximum possible for comparison asd above. In above case ‘ef’ is put at

last to show maximum possible case.

Here number of loops=

Zero to (length of in find string- length o two find string)

i.e. 5 =1+(6-2)

5=5 proved.

Here

for(j=i,k=0;j<(l2+i);j++){

total count is equal to l2, so this loop has number of loops equal to l2 as we know we need to

compare l2 with other so l2 is importantly taken under consider.

Here,

temp[k]=input[j];

k++;

}

for each new loop of I, k startsfrom zero i.e. new comparison starts up to the number of characters

equal to l2 but for each loop of j,k increases trhat in new position of temp[] variable is fulfilled and

after its length becomes equal to length of l2 trhen comparison is done.

The value of j starts from mi because after each comparisons done at last, there is no necessity to

compare for those characters already compared so the value of j starts from previous value of I i.e.

new characters are compared each time and in the other hand the complete comparison for all

characters is possible and termination is also possible.

//searching a character or string in a sentecnce

#include<stdio.h>

#include<conio.h>

#include<string.h>



101

int main() {

char input[100],search[100];

printf("enter a sentence\t");

gets(input);

printf("enter specific word\t");

gets(search);

int l1=0,l2=0,i,j,k;

l1=strlen(input);

l2=strlen(search);

char temp[100];

for(i=0;i<=(l1-l2);i++){ // the maximum possible condition to find word

for(j=i,k=0;j<(l2+i);j++){

temp[k]=input[j];

k++;

}

temp[k]='\0'; //compulsory else the blank spaces may make obstacle and

temp is not a complete string as'\0' is last character of all string

if(strcmp(temp,search)==0) {

printf("found");

break;

}

}

getch();

return 0;

}



102

A DETAIL OF STRINGS

Finding a sequence of words

// searching whole sentence

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main(){

char check[50]="";

int i=0,j=0,nooffound=0;

char name[100]="NEPAL A HEAVEN";

printf("ENTER");

scanf("%[^\n]",&check);

for(i=0;i<strlen(name);i++) {

if(name[i]==check[j]){

nooffound++;

j++;

continue; //this may be omitted

}

}

if(nooffound==strlen(name)){

printf("all words matched");

}

else if(nooffound==0) printf("no matching with a character even");

else printf("some characters or words are matched but not all");

getch();

}

Finding a sequence of words any

//searching sequential characters

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main(){

char check[50];

int i,nooffound=0;

char name[100];

int max=0;

int count=0;

int temp=0;

printf("ENTER a sentence\n");

scanf("%[^\n]",&name);

printf("ENTER\n");

scanf("%s",&check);




103

if(name[temp]==check[count]){

printf("I=%d",i);

nooffound++;

if(nooffound>max) {

max=nooffound;

}

count++;

temp++;

}

else {

if(nooffound>max) {

max=nooffound;

}

nooffound=0;

if(count>0) {

count=0;

}

temp++;

}

if(count==strlen(check)) {

count=0;

}

if(nooffound==strlen(name)){

printf("exactly found");

if(getch()) exit(0);

}

}

printf("Not foundexactly");

printf("\n[%d]characters are in sequence",max);

getch();

return 0;

}

Finding a sequence of words (working)

//searching sequential characters

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main(){

char check[50];

int i,nooffound=0;

int max=0;

int count=0;

int temp=0;

char name[100]="nepal is a hindu country";

printf("ENTER\n");



104

scanf("%[^\n]",&check);


printf("\ncount=[%d]",count);

printf("\ncheck=[%c] and name=[%c]",check[count],name[temp]);

if(name[temp]==check[count]){

if(nooffound>max) {

max=nooffound;

}

nooffound++;

printf("NO OF FOUND%d",nooffound);

count++;

temp++;

}

else {

if(nooffound>max) {

max=nooffound;

}

if(count>0) {

count=0;

}

temp++;

nooffound=0;

}

if(count==strlen(check)) {

count=0;

}

if(nooffound==strlen(check)){

printf("exactly found");


}

}

printf("Not foundexactly");

printf("\n[%d]characters are in sequence",max);

getch();

return 0;

}

Finding a word in a sentence

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main(){

int i;

int a=-1;

char incheck[50];

char name[50]="My country is Nepal , a heaven";

char check[50]="";

clrscr();



105

printf("My country is Nepal , a heaven\n\n");

printf("ENTER A WORD\n");

scanf("%s",check);


a++;

incheck[a]=name[i];

if((strncmp(incheck,check,(strlen(check)))==0)) {

printf("found\nInput any thing to exit");

if(getch())exit(0);

}

else if(name[i]==' '){

strcpy(incheck,"");

a=-1;

}

}


getch();

return 0;

}

Finding a word in a sentence (working)

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main(){

int i;

int a=-1;

char incheck[50];

char name[50]="My country Nepal is heaven";

char check[50]="";

printf("My country Nepal is heaven\n");


scanf("%s",check);


a++;

incheck[a]=name[i];

printf("\ncheck=[%s] and incheck=[%s]\n",check,incheck);



if(getch())exit(0);

}


printf("above space");

printf("incheeeeeeeeeeeeeeek=%s\n",incheck);

strcpy(incheck,"");



106

a=-1;

}

}


getch();

return 0;

}

Finding a word in a sentence (input from user)

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

int main(){

int i;

int a=-1;

char incheck[50]="";

char name[50]="";

char check[50]="";

printf("Enter A Long Sentence\n");



scanf("%s",&check);


a++;

incheck[a]=name[i];



if(getch())exit(0);

}


strcpy(incheck,"");

a=-1;

}

}


getch();

return (0);

}



107

OPERATORS

The numeric values used in any expression or used like -2,3,5.0,10.67,10e-2 are operands but the

signs or symbols used to operate between operands are called operators. Operators are various

types listed below:-

TABLE

Rank Operator Associativity

1 (), [] L to R

2 + (unary), -(unary),++,--,! ,*(indirection) R to L

3 *,%,/ L to R

4 +,- L to R

5 <<,>>(shifts) L to R

6 <,<=,>,>= L to R

7 ==,!= L to R

8 && L to R

9 || L to R

10 ?:(conditional or tertiary) R to L

11 =,*=,/=,%=,+=,-=,&=(assignment operators) R to L

12 . (dot) ,(comma) L to R

The rank list shows the priority of which operators to take execution at first or second …….and

similarly so on. The inner () has highest priority than gradually surrounding ().

Inside the same rank, the table shows direction of associativity - L to R means left to right and R to L

means right to left. This direction shows how the calculation’s execution takes place. If in any

expressionthere is more than on operators of a same rank, then the associativity directs the flow of

execution. The operators directed according to associativity come into execution as an when it is

encounted.

For e.g.

100+20/(3+5)*(6+2)

Here () has 1KLpriority

/,* has 2G=priority

+ has 3@=priority

So at 1KL() must be calculated but there are two (3+5) and (3+2) expressions but as we know () has L

to R associativity, (3+5) must be first calculated and (6+2) second resulting,

=100+20/8*(6+2)

=100+20/8*8

Now we come for 2G=priority /,*

It has also left to right associativity, so

=100+2*8 // 20/8 gives 2 as it is integer division (is that takes operates between 2 integer

operands) that truncates decimal part



108

=100+16

Now + is operated which has 3@=priority,

=116

Few more examples,

20%8*30(L to R associativity)

= 4*30

= 120

HOW ERRORS MAY OCCUR?

ERROR CALCULATION

3.5+12%6*3

=3.5+12%18

=3.5+12

=15.5

Here %,* has L R associativity but in our calculation we didin’t consider it and we did R to L

asscociativity knowingly or unknowingly, this is our mistake.

In numeric calculations, one mistake is sufficient to bring wrong answer.

CORRECT CALCULATION

3.5+12%6*3

=3.5+0*3

=3.5+0

=3.5

Comparing the underlined parts of these two calculations we see that one gives 12 but another 0,

this shows the actual difference which occurred due to incorrect associativity. Hence the above two

examples illustrates the vast differences due to small errors.

For leap year

A leap year is a year if it is divisible by 4 but not by 100, except that years divisible by 400 are leap

years.



109

if((year%4==0 && year%100!=0) || year%400==0) printf("leap year");

else printf("Not leap year");

a%b; here % is modulo division that gives remainder when a is divided by b, but a and b must be int

type as % cannot accept float datas.

Let us give year=20 then

if((20%4==0 && 20%10!=0) || 20%400==0)

Here the inner () has 1KL priority. Inside this inner parenthesis (), % has 1KL priority but there are two

% signs so by L-R associativity of %we get

�1KL step =if((0==0 && 20%10!=0) || 20%400==0)

�2G= step =if((0==0 && 0!=0) || 20%400==0)

Now, ==, != has priority with L-R associativity

�3@= step =if((True && 0!=0) || 20%400==0)

�4L` step =if((True && False) || 20%400==0)

�5L` step =if((False) || 20%400==0)

Because True and False gives False refer Truth Table

=if(False || 20%400==0)

In above case, we did all calculations of inner () so outer is now to be taken consideration.

Here, % has 1KL priority ,== 2G= priority and || 3@= priority.

So,

�6L` step =if(False || 0==0)

�7L` step =if(False || True)

�8L` step =if(True)

So 20 is the leap year, we did it in 8 steps.

POST/PRE INCREMENT/DECREMENT OPERATORS



110

Prefix(before the variable: ++a) and postfix(after the variable: a++)

++n increments n before its value is used

n++ increments n after its value has been used.

LINE POST PRE

1 x=5; //x=5 x=5; //x=5

2 x++; //x=5 ++x; //x=6

3 x; //x=6 x; //x=6

In 2G= line of pre-, first x is incremented by 1 i.e. x becomes 6 and this value of x is used however in 2G= line of post- the value of x=5 is used then only after its one time use, the x is incremented by 1

i.e. x becomes now 6 which is shown by line 3 of post, but x becomes 6 already from line 2 of pre-.

In post

a++ or a- -

In this particular line value of ‘a’ is original’s but the next time or may be on next line the value is

initially incremented or decremented before than under-consideration of ‘a’ or statements or

expressions including ‘a’ or including its parts.

FOR-BLOCK’S EXECUTION FLOW

For block

Initialize a value (let say i)

Check condition (if true then follow statement else terminate for-loop)

Statement

Statement

Statement

.

.

.

.

.

Update value controlling for loop(here it is i)

(actually there are two controlling methods one is counter-controlling and the next is sentinel-

controlling loop, counter controlling is definite but in sentinel it is not definite and the termination is

determined by some undefinite terms like may be user’s input)

Illustration of pre/post operators in detail

for(i=1;i<+5;i++)

print("%u",i);

and



111

for(i=1;i<=5;++i)

print("%u",i); // here u means unsigned int

These two for-expressions use post and pre increments of i but does not hamper the statements

inside, it holds. It is because both post and pre are declared in one line and its use like printing i is

declared in next line or a next statement, so the output are same. But let us look an example:-

for(i=1;i<+5;a=i++){

print("%d",i);

print("%d",a);

}

and

for(i=1;i<+5;a=++i){

print("%d",i);

print("%d",a);

}

Here in a=i++, ++ has first priority than = so, first is 1++, which means that i has 1 value but in next

statement that using i, there i will be 2, but here a=i++ takes the instant value of i in the line i.e.

before increment as it is for only the next statement using i, where i will be incremented. So a=1 as it

is, but in second example, ‘a’ will be 2.

We are not going to deal with these post/pre fix operators in detail here however we will clear them

when we encounter them in coming parts.

ASSIGNMENT OPERATORS

i=i+2=5;

here = has R to L so, i+2=5 �1KL step

i=i+2 �2G= step

but from 1KL step i+2=5

so, from �2G= step i=5



112

OUTPUTS

NOTE: Before looking outputs, please once more revise operators-section.

HINTS TO pu2008fall 3a)

x=++x+y;//++x has first priority so,++3+4=4+4=8

z*=(x+y) //z=z*(x+y)

3b)

#include<conio.h>

#include<stdio.h>

void main() {

int x=0,i=0;

for(i=1;i<10;++i)

{

if(i%2==1)

x+=i;

else

{

x--;

printf("%d",x);

break;

}

}

printf("\nx=%d",x);

getch();

}

i Condition

i%2==1

Result Operations Remarks

1 1%2==1 True x=x+i

X=0+1

2 2%2 False x- - X=1—

print 0

Here in the line x--, the value of x is original that is 1, but in the following line print x, the value of x is

0.

Now the ‘break;’ statement breaks the for-loop and the value of “x=0” is printed.


PU 2006 SPRING2 a)

#include<stdio.h>

#include<conio.h>



113

int a = 100, b = 200;

int funct1 (int c);

main ( )

{

int count, c;

for (count = 1; count <= 5; ++count) {

c= 4 * count * count;

printf ( " %d ", funct1 ( c ) );

}

getch();

}

funct1 ( int x)

{

int c;

c = (x < 50) ? (a + x) : (b - x) ;

return( c );

}

We can use table under appropriate column names to make easier to calculation like below:

count c=4*count*count condition x<50 result operations

1 c=4*1*1=4 4<50 TRUE c=a+x=100+4=104

print 104

2 c=4*2*2=16 16<50 TRUE c=a+x=100+16=116

print 116

3 c=4*3*3=36 36<50 TRUE c=a+x=100+36=136

print 136



114

4 c=4*4*4=64 64<50 FALSE c=b-x=200-64=136

print 136

5 c=4*5*5=100 100<50 FALSE c=b-x=200-100=100

print 100

2 b)

#include<stdio.h>

#include<conio.h>

main( )

{

int i=0,x=0;

for(i=1;i<10;++i )

{

if (i%2==1)

x+=i;

else

x--;

printf("%d", x);

continue;

}

printf ( " \nx = %d ", x );

getch();

}

i Condition

i%2==1

Result Calculation first Calculation second

1 1%2==1 true x=x+i

x=0+1 i.e. 1

print 1

continue loop

// continue; statement simply continues the for-loop in this case however it may make greater

difference for others where the statements below continue-statement but inside for-loop are just

aborted.



115

2 2%2==1 false x--=1-- print 0

3 3%2==1 true x=x+i=0+3=3 print 3

4 4%2==1 false x--=3-- print 2

5 5%2==1 true x=x+i=2+5=7 print 7

6 2%2==1 false x--=7-- print 6

7 7%2==1 true x=x+i=6+7=13 print 13

8 2%2==1 false x--=13-- print 12

9 3%2==1 true x=x+i=12+9=21 print 21

2) c

#include <stdio.h>

#include<conio.h>

#define ROWS 3

# define COLUMNS 4

int z [ROWS][COLUMNS] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};

main ( )

{

int a, b, c;

for ( a = 0; a < ROWS; ++a)

{

c = 999;

for ( b = 0; b < COLUMNS; ++b )

if ( z[a][b] < c ) c = z[a][b];

printf ( " %d ", c );

}

getch();

}

While checking 999 with all numbers, no new number is > than 999 so c is 999 and 999 is thus

printed.



116

PU2009SPRING

2 b) important

#include<stdio.h>

#include<conio.h>

void main()

{

int sum,a,k,j;

sum=0;

a=12345;

while (a!=0)

{

j=a%10;

printf("%d",j);

if(j%2==0)

{

sum=sum+j;

}

a=a/10;

}

printf("sum=%d",sum);

getch();

}

The exact question inquestion paper is mistake as while(a!=0) while(12345!=0) never becomes

false and no termination so we keep a=a/10 and the purpose of program seems that it sums all even

digits of ‘a’ and prints sum.

Condition

(a!=0)

Result (T/F) Computations

(j=a%10)

and print j

Condition

(j%2==0)

Result Calculation Valve of a

(a=a/10)

12345!=0 T j=12345%10=5

print 5

5%2==0 False - a=12345/

10=1234

1234!=0 T j=1234%10=4

print 4

4%2==0 True sum=0+4=4 a=1234/1

0=123

123!=0 T j=123%10=3

print 3

3%2==0 False - a=123/10

=12

12!=0 T j=12%10=2

print 2

2%2==0 True sum=4+2=6 a=12/10=

1

1!=0 T j=1%10=1

print

1%2==0 False - a=1/10=0

0!=0 False (------end of while-loop------)

print sum =6



117

3 a) very important

#include<stdio.h>

#include<conio.h>

void main()

{

int x=0, i=0;

for (i=1; i<10; i++){

if (i%2==0)

x += i;

else

{

x -- ;

printf ("%d\t", x);

}

}

printf("\nx=%d", ++x);

getch();

}

i Condition

(i%2==0)

Result Calculation Update i of for-

loop

(i++)

1 1%2==0 False x=0- -

print -1

i=1++

(see how 1++

changes to 2 in

next statement)

2 2%2==0 T x=x+i=-1+2=1 i=2++

3 3%2==0 F x=1- -

print 0

i=3++

4 4%2==0 T x=x+i=0+4=4

i=4++

5 5%2==0 F x=4--

print 3

i=5++

6 6%2==0 T x=x+i=3+6=9

i=6++

7 7%2==0 F x=9--

print 8

i=7++

8 8%2==0 T x=x+i=8+8=16

i=8++

9 9%2==0 F x=16--

print 15

i=9++



118

i=10 but i<10 is only permissible for for-loop, so loop terminates.

print x=16 //++15 is 16

PU2011FALL

3 a)

#include<stdio.h>

#include<conio.h>

void main()

{

int i=5,j=4,k=9;

i=(i+k)/3+k%(j+i)+j*k%i/2;

printf("%d",i);

getch();

}

i=(5+9)/3+9%(4+5)+4*9%5/2

here () has highest priority and follows L to R associativity. So,

i=14/3+9%(4+5)+4*9%5/2 (1�� step)

i=14/3+9%9+4*9%5/2 (2�� !.a

now, %,*,/ has high precedence with L to R associativity, so

i=4+9%9+4*9%5/2 �3�� !.a

i=4+0+4*9%5/2 �4�6 !.a

i=4+0+36%5/2 (5�6 !.a

i=4+0+1/2 (6�6 !.a

i=4+0+0 (7�6 !.a

now + has high precendence than =,so

i=4+0 (8�6 !.a

i=4 (9�6 !.a

now the value of i is 4, print i //4 prints out



119

3 b)

#include<stdio.h>

#include<conio.h>

void main()

{

int x=16,y=18,z;

y+=x++;

z=++x+y++;

x=x+y+z--;

printf("\n x=%d\ny=%d\nz=%d",x,y,z);

getch();

}

y+=x++;

her, ++ has high precedence than assignment operator +=. So, at first x++ is calculated. 16++ means x

has 16 value o the line but 17 on next line.

So, y=y+16++

=18+16++

=34

In next line, ++ has R to L associativity so , at first y++ is calculated, then ++x, then + is done and

finally assigned to z.

z=++x+34++ (1�� step)

=++17+34++ (2�� step)

=18+34++

=52

z=52

In 3�� line

x=18+35+52--

=18+35+52

=105

Line x y z

Value

before

calculation

of line

Value after

calculation

of line

Value

before

calculation

of line

Value after

calculation of line

Value before

calculation of

line

Value after

calculation

of line

1 16 16++ 18 18+16++=18+16=34 Not

defined(garbage

value)

Not

defined( it

gives error



120

like no

value

declared

when

printed)

2 17 ++17=18 34 34++ - 52

3 15 105 35 35 52 52--

4 105 105 35 35 51 51

Print x=105, y=35, z=51

Hints: Read each line of statement separately and make a general table like above which includes

values of x,y,z in various lines and sub-divide the values into two parts- one stores value before the

calculation of the line is done and the next stores value after the calculation of the line has been

done. In order to do calculation for a particular line we need to take the value from “value before

calculation of line” and do our all calculations of the line. In calculation of a line, a value may go

under various sort of changes it doesn’t matter to us and we have to remember a thing that a

variable has only one address so a value changed from any where or at any time automatically

changes the value where its old values are erased without any delay so one by one step method

makes everything clear.

For e.g. a=5

a++,b=a- -

Value of a before calculation, a=5

In calculation, 5++ but later changed to 6 on next using of ‘a’

Then after calculation a- -means 6- - on next line it is a=5 sometime a variable’s value may be

changed more than 1 time abruptly within a single statement or line.

Let’s return to our lesson, after the calculation we need to fill the “value after calculation of line”

with the new arising values but we should be careful that I post operators write the symbol++ or - -

as it is so we change it into its real form by initialization ofnext line.

For e.g. the x is 16++ of line 1 changes to x is 17 in line2. So this table-work must go side-by-side with

the line-by-line calculation.

PU2009FALL 3 a)

#include<stdio.h>

#include<conio.h>

void main()

{

int m = 100, n = 50;

printf ("%d\n", 10+m++); //1^st line

printf ("%d\n", 10+n--); //2^nd line



121

printf ("%d\n", 10+--n); //3^rd line

getch();

}

1�� line 10+#77

10+10077 // 100 for this line and 101 for next

print 110

2�� line 10+,55

10+5055 // 50 for this line and 49 for next

print 60

3�� line 10+55n

10+5548

print 58

Line number Value of m Value of n Printing on

line Before After Before After

1 100 10077 50 50 10+

10077=110

2 101 101 50 5055 1+

5055=60

3 101 101 49 5549

=48

10+48=58

Neglect unnecessary values in the table for the case on line 2 and 3 values of ‘m’ and value of ‘n’ on

line 1. But in long statements, this unnecessary tabulation may give a great relief to us.

3 b)

#include<stdio.h>

#include<conio.h>

void main()



122

{

char string [ ] = "HELLO WORLD";

int m;

for (m=0;string[m]!='\0';m++)

{

if (m%2==0) printf ("%c",string[m]);

}

getch();

}

m Condition of for-

loop

string[m]!= ‘\0’

Condition

(m%2==0)

Result Calculation Update of m

(m++)

0 True 0%2==0 T print H 0++

1 T 1%2==0 F - 1++

2 T 2%2==0 T print L 2++

3 T 3%2==0 F - 3++

4 T 4%2==0 T print O 4++

5 T 5%2==0 F - 5++

6 T 6%2==0 T print W 6++

7 T 7%2==0 F - 7++

8 T 8%2==0 T print R 8++

9 T 9%2==0 F - 9++

10 T 10%2==0 T print D 10++

11 F (-------for-loop terminates--------)

Here we are checking the condition of for-loop in order to demonstrate how character array works.

OUTPUT: HLOWRD

PU2007FALL 2 a)

#include<stdio.h>

#include<conio.h>

void main()

{

int x =10,a =10,b=0, y =11, z =12;

x+=++a-b++; // line

x-=++z%--b; // line

x*=++a+b++*z-- ; // line

printf("a=%d\tb=%d\tx=%d\ty=%d\tz=%d",a,b,x,y,z);

getch();

}



123

Without assigning a variable a value and using it makes error but it doesn’t hamper for assignment

so we have supposed b=0 in the question. But even after that something divide by b (zero) gives

complilation error so the program gac’t run. However here we are going to rewad the process of

calculation by assuming error as correct.

line ++ has priority and R L associativity

x=x+(++a-b++) (first step)

x+(++10)-0 ( second step)

x+11-0

x=x+11 ( step)

x=10+11 =21

line 2

x-=++z%- -b

x-=++z%- -1 (first step)

x-=++12%0// here int divide by zero may give error

x- =13%0 (second step)

x-=13 ( third step)

x=21-13=8

write the actual values instead of variable only when the particular calculation is to be done else it

will confuse you.

Line 3

x*=++a+b++*z- -

x*=++a+b++*13- - (first step)

x*=++a+0++*13

x*=++a+0*13 ( second step)

x*=++11+0*13

x*=12+0*13 (third step)

x*=12+0 ( fourthstep)

x*=12 ( fifthstep)

x=x*12 ( sixth step)

x=8*12=96

Line x y z a B

before after before after before after before after before after

1 10 21 11 11 12 12 10 ++10=11 0 0++

2 21 8 11 11 12 13 11 11 1 0

3 8 96 11 11 13 11 ++11=12 0 0++

Here, now on the particular line after execution

x=96, y=11 (never changed in any line), z=(i.e. 13 on line but changes 12 on line which is not defined

in program), a=11,b= (i.e. 0 on line but 1 on coming line).



124

About answer of c; it is undeclared variable so it gives error as undeclared variable when access ‘c’ in

program.

2 b)

#include<stdio.h>

#include<conio.h>

void main()

{

int k, a;

a = 10;

for(k=0;k<10;k++)

{

if(k%2==0)

a = a+k;

else

a = a-1;

printf("a=%d\n",a);

}

printf("\n the value of a = %d", a);

getch();

}

k Condition

(K%2==0)

Result Calculation print a

0 0%2==0 T a=a+k=10+0=10 print 10

1 1%2==0 F a=a-1=10-1=9 print 9

2 2%2==0 T a=a+k=9+2=11 print 11

3 3%2==0 F a=a-1=11-1=10 print 10

4 4%2==0 T a=a+k=10+4=14 print 14

5 5%2==0 F a=a-1=14-1=13 print 13

6 6%2==0 T a=a+k=13+6=19 print 19

7 7%2==0 F a=a-1=19-1=18 print 18

8 8%2==0 T a=a+k=18+8=26 print 26

9 9%2==0 F a=a-1=26-1=25 print 25

print a=25



125

2012FALL 2b i) Very important

#include<stdio.h>

#include<conio.h>

void main()

{

int x=10,y=20,z=5,i;

i=x*(++y+z++)%3+y/x*2-5;

x+=z;

y-=z;

x--;

z+=10;

z--;

printf("x=%d\ny=%d\nz=%d",x,y,z);

printf("\ni=%d",i);

getch();

}

1�� line () has high priority in which ++ has 1�� play with R to L associativity then outside () for

*,/,% with L to R associativity then +,- eith L to R associativity,

Here steps are written when actual calculation is done excluding actual value assignment.

1�� line

i=x*(++y+z++)%3+y/x*2-5

= x*(++y+5++)%3+y/x*2-5 1��step

=x*(++20+5)%3+y/x*2-5

= x*(21+5)%3+y/x*2-5 2��step

= x*26%3+y/x*2-5 3��step

= 10*26%3+y/x*2-5

=260%3+y/x*2-5 4�6step

=2+21/10*2-5 5�6step



Note:- what happen if these way z/x instead earlier said that a post operator’s actual value comes

into play either in next line if there is no further use of that variable in that line or if there is use in

same line that the actual and final value must come into pla

must write 6 for the case.

=2+2*2-5 6�6step

=2+4-5 7�6step

=6-5 8�6step

=1 9�6step

So i =1

TABLE

Line x

Before

calculation of

line

After

calculation of

line

1 10 10

2 10 16

3 16 16

4 16 16- -

5 15 15

6 15 15

7 10 No any

calculation

2�� line x+=z i.e. x=x+z=10+6=16

3�� line y-=z i.e. y=y-z=21-6=15

4�6 line x- - i.e. 16- -

5�6 line z+=10 i.e. z=z+10=6+10=16

6�6 line z- - i.e. 16- -

Print x= 15 y=15 z=15 i=1


happen if these way z/x instead earlier said that a post operator’s actual value comes


same line that the actual and final value must come into play so instead of 5++ in Table for z, we

y z

calculation of

Before

calculation of

line

After

calculation of

line

Before

calculation of

line

20 ++20=21 5

21 21 6

21 21-6=15 6

15 15 6

15 15 6

15 15 16

calculation

15 No any

calculation

15

line x+=z i.e. x=x+z=10+6=16

6=15

line z+=10 i.e. z=z+10=6+10=16

126

happen if these way z/x instead earlier said that a post operator’s actual value comes


y so instead of 5++ in Table for z, we

calculation of

After

calculation of

line

5++

6

6

6

16

16- -

No any

calculation



127

PU 2011 SPRING Ternary Operator’s nested use

#include<stdio.h>

#include<conio.h>

void main() {

int i=-4,j,num=0;

j=(num>5?(num<=10?100:20):500);

printf("%d",num);

printf("%d",j);

getch();

}

Output:

0500

Here inner () is operated at first in ternary operator

num<=10

0<=10 (True)

So, j=(num>5?100:500)

As the inner condition condition becomes true so 100 is taken but remember it is not assigned to

num, num still has 0 value.

Now, num>5

0>5 (False)

So, j=500. While printing num=0 and j=500 so 0500is printed out.

PU2006FALL 3

Print out combinations of 1, 2, 3 using for-loop

#include<stdio.h>

#include<conio.h>

void main(){

int i,j,k,num,flag=1;

for(i=1;i<=3;i++) {

flag=1;

num=i;

for(j=1;j<=3;j++) {

if(j==i) continue;

num=num*10+j;



128

for(k=1;k<=3;k++) {

if(k==i || k==j) continue;

num=num*10+k;

printf("%d\n",num);

flag=0; break;

}

if(flag==0) break;

}

}

getch();

}

PUFALL2003

2 b)

Read temperature of a month of 10 cities and print out average of ten cities using array and for-loop

#include<stdio.h>

#include<conio.h>

void main(){

int i,j,sum=0;

char city[10][25];

int tmp[10][30];

for(i=0;i<10;i++) {

printf("Enter name of city");

scanf("%s",&city[i]);

}

for(i=0;i<10;i++) {

for(j=0;j<30;j++) {

printf("Enter temperature of %s for %d day",city[i],j+1);

scanf("%d",&tmp[i][j]);

}

}

for(i=0;i<10;i++) {

sum=0; //for each value of ‘i’ i.e. for eachnew city sum must starts from zero

for(j=0;j<30;j++) {

sum=sum+tmp[i][j];

}

printf("average of %s city is %d",city[i],sum/30);

}

getch();

}



129

SECOND WAY OF SOLVING

#include<stdio.h>

#include<conio.h>

void main(){

int i,j,avg[10];

char city[10][25];

int tmp[10][30];

for(i=0;i<10;i++) {

printf("Enter name of city");

scanf("%s",&city[i]);

}

for(i=0;i<10;i++) {

avg[i]=0;

for(j=0;j<30;j++) {

printf("Enter temperature of %s for %d day",city[i],j+1);

scanf("%d",&tmp[i][j]);

avg[i]=avg[i]+(tmp[i][j]/30);

}

}

for(i=0;i<10;i++) {

printf("average of %s city is %d",city[i],avg[i]);

}

getch();

}

Here average is calculated directly as the sum of such values which has got ratio (or fraction) to the

total number of species i.e. 30

For e.g, if there are data 10, 50,60,5 then its average can be calculated as

Avg=(10+50+60+5)/4=31.25 (in I CASE)

But in second case it is calculated as

Avg=10

4+50

4+60

4+5

4=31.25

As here 4 represent the total quantity of data, formerly in program it was 30



130

POINTERS

Pointer declaration in arrays

1-D array:

Declaration *a[3]={1,2,3};

To access *a[i]; such that i running from 0 to 2

2-D array:

Declaration *(a+i)+j; or (*a)[]; or *a[][];

// or to pass value by address or pointer

To access *(*(a+i)+j); or *(a[i]+j); // to use value

int x,*p,y;

x=5;

p=&x;

y=*p; //accessing x through p-pointer variable

*p=50; //accessing 50 to x through pointer p

In 1-D,

*(x+i) represents 1-D array x[i] and *(*(a+i)+j) represents a[i][j]

In p[i][j]

p pointer to 1�� row

p+i pointer to ��6 row



131

*(p+i) pointer to 1�� element in ��6row

*(p+i)+j pointer to b�6 element(column) in ��6row

*(*(p+i)+j) value of ��6 row and b�6 column

PROGRAM DEMONSTRATING POINTER VARIABLE’S RESEMBLENSE WITH SPECIAL ARRAY i.e.

CHARACTER ARRAY (STRING)

N E P A L \0

Address 54 59

(name) (ptr)

#include<stdio.h>

#include<conio.h>

void main(){

char *name;

char *ptr;

name="NEPAL";

ptr=name;

while(*ptr!='\0') {

printf("%c is stored at address %d\n",*ptr,ptr);

ptr++;

}

printf("\n length=%d\n",ptr-name);

getch();

}

DESCRIPTION

The pointer ptr points the name variable so ptr gives address of each characters of name variable i.e.

“NEPAL” and similarly *ptr access each characters.



132

After while termination, the ptr has gained one more succeeding address after “L” of NEPAL i.e. of

“\0” so its subtraction with name which has address of 1��character of name variable i.e. “N” of

“NEPAL”gives length of “NEPAL”.

USING VOID TYPE POINTER AS REQUIRED

void* v; //a void pointer

int a =50;

v=&a;

printf("%d",*(int*)v); //prints 50

//here '(int*)v' declare v as int type pointer

Demonstrating the 2D pointers in 2d array

#include<stdio.h>

#include<conio.h>

int main() {

int*matrix[2][2];

int i,j;

printf("first matrix");

for(i=0;i<2;i++)

for(j=0;j<2;j++) scanf("%d",*(matrix+i)+j);

//printf("\n%d",*(matrixa+i)+j);

for(i=0;i<2;i++)

for(j=0;j<2;j++)

printf("output:%d",*(*(matrix+i)+j));

getch();

return 1;

}

CALL BY ADDRESS/PASS BY POINTER

int x=5;

cahnge(&x); // call by reference or address

printf("%d",x); // x=15

change(int* p) {

/*here in int* p, * has R-L associativity.

So, p is a variable of type pointer as p has priority than * and of type int.

Here the input as argument is taken by p not *p, *p shows the p is pointer type alse we can say int*

is one thing and p another, int* showing type of value taken by p and type of p-variable i.e. int and

(pointer) respectively.

*/

*p=*p+10;



133

}

POINTERS IN STRUCTURE

struct student *ss;

to assign or access value we use ‘(*ss).id’ //* has R-L associativity but we need L-R associativity so ()

used

or we use ‘ss->id’

PASSING WHOLE STRUCTURE TO FUNCTION

function_name(struct_name obj); //calling

and we can directly accesss elements of structure inside function

PU 2006 SPRING 5a

Meaning of some pointer expressions:

Qn. Explain the meaning of the following declaration:

i) int *p;

ii) int *p[10];

iii) int (*p)[10];

iv) int *p(void);

v) int (*p)(char*a);

Answer:

Meaning of i):

It simply declares a variable ‘p’ which is pointer type that means it stores address of some other

variable but this pointer type is int so , it stores only address of int type some other variable.

Meaning of ii) and iii):

*has lower precedence than [], so *p[10] declare p has an array of 10 pointers but (*p)[10]

declare p as an pointer to an array of 10 elements.

{*has R-L}and{(),[] has L-R}



134

Meaning of iv):

POINTER TO FUNCTIONS:

iv) int *p(void); declare p as a function (taking void means nothing as argument) returning a

pointer type int value.

It is because () has high priority and L R associativity i.e. p(void) p is a function wioth no

argument and similarly * has R to L associatrivity in addition to 2�� priority so

int*p(void) ;

is a function returning pointer type.

Meaning of v):

int (*p)(char*a);

The 1�� priority () has L to R associativity. So, (*p) is taken 1�� than (char*a) i.e. p is a

pointer to a function taking argument as pointer type vartiable-a and this function returns

int value.

2003 fall 4 a) Write a program with user defined function using pointers to convert all the upper-

case letters to lower-case and vice versa in a string given by the user.

For example: The C pRogramming.

tHE c PrOGRAMMING.

#include<stdio.h>

#include<conio.h>

#include<string.h>

int main() {

void converter(char*);

char string[50];

printf("enter a string\t");

scanf("%[^\n]",&string);

converter(&string);

getch();

return 0;

}

void converter(char* string){

int i;

for(i=0;i<strlen(string);i++){

int a=string[i];

if(a>=65 && a<=90) { // if we look the ASCII value table this range is for Upper case

a=a+32; // the ASCII value difference betweenlower and upper

is32

}

else if(a>=97 && a<=122) { // similarly this range is for Lower case

a=a-32;



135

}

printf("%c",a);

}

}

SOME CONCEPTUAL TERMS

RETURNING MULTIPLE VALUES BY FUNCTION

Show how pointers can be used to return more than one values from a function

(PU2003FALL4 a)]

A function is a block that takes arguments (char, int, float of normal type or pointer type) and does

either some sort of calculations or other processes and may either return a value or return no value

as a direct visible output. So realizing in a direct method or by just using simple ordinary variables, a

function can return at most one value, but by using pointer variables, a function can return multiple

values. These output parameters are obtained by &(address operator) and indirection operator(*).

For e.g.

#include<stdio.h>

#include<conio.h>

void calculation(int x, int y, int* sum, int* sub) {

*sum=x+y;

*sub=x-y;

}

void main() {

int x=5, y=2, a, s;

calculation(x,y,&a,&s);

printf("Sum=%d\t diff=%d",a,s);

getch();

}



136

Here as argument two pointer variables are passed in addition to two simple variables. The sum and

diff is assigned to address of ‘a’ and‘s’ or to pointer variable’s pointing address. Now from the main

method these values are accessed. So the function does calculations and stores values in addresses

by help of pointers.

A function directly returns at most only one variable but can choose what value to be returned

among many depending upon conditions- meeting.

PROGRAM DOCUMENTATION

It means remarkations used to give information or declaration of some lines. Two types-single line

and multiline denoted by //…….and /*………*/. This documentation increases the readability sand in

execution of program, these statements arte left and the flow of execution rejects it and directly

goes to next line so it doesn’t hamper the speed of execution of program. So it is advantageous.

Documentation makes easy to understand a program even for beginning users and also to

programmer himself if for a big program, he might lose concentration after long programming or

recovering the program after a long time.

IMPORTANT CASTING MECHANISM

#include<stdio.h>

#include<conio.h>

void main() {

int a=20,b=6;

float r1,r2;

r1=a/b;

printf("%f",r1);//3.000000

r2=(float)a/b;

printf("\n%f",r2);//3.333333

printf("\n%f",(float)(a/b));//3.000000

getch();

}

Here on first printing float is in top hierarchy than int so int value 3(because 20/6=3 in int integer

division)is converted into float by ‘r1=a/b;’ statement, but in second printing the ‘a’ is converted into

float by ‘=(float)a/b;’ and here () has precedence than / and the result is in float type, however in

third printing ‘(float)(a/b)’, first up all a/b is done which is 3 then only it is changed into float so

printing 3.00000.

VARIABLES

Automatic variables are those local variables (variables accessible within only a specified block)

declared inside functions so that it is only accessible within function, it make up its own new address



137

so that it does not alter the variable of main method even if main method has the same name

variable and at last, these variables are generated only after function is called and their lifetime is

limited until the end of execution of the same function for the same particular call of function i.e.

when we call a function for a number of times then the local variable are evolved and terminated for

the same number of times, and each time creating new space, new value (may be some

initialization) where old values have no existence.

But if we want to use the old values of local variables then static type local variables must be

declared, i.e. in static type local variables within between many calls of a same function, the value of

a variable remains still in the sense that it exits but, changes in the sense that it may not hold a value

for the all time in a specified position i.e. in automatic variable at initialization a variable holds a

same value for all calls but this mightn’t be in static. Here the variable is generated when a first call is

given to function and it only terminates on whole execution of program.

Global variables are accessible, changeable and assignable from any part of a program- main method

or any sub-module, module, block, etc.

CALL BY VALUE VS CALL BY REFERENCE

What actually is a function? A function is a block that does some operation. It may do two types of

operation and by two different ways. One operation sis the internal operation which plays role for

inner calculations and the next operation adds some feature in addition to former operation i.e. the

operation is visible to where the function is called or the function returns some value. This value

returned is very much co-related to the way by which the function is operated and we know a

function is operated by two ways- one is by passing some values as input in order to process it or

operate it whereas the other is passing no value at all. But now, we are concerned with methods of

passing values and memory space.

In call by value, the actual arguments we pass in function are copied to some local variables of the

function. As we know local variables specify their own addresses or locations so if we now make

some changes to the variable then we make change for local variable not for the real variables

passed from main method.

int a=5,b=2;

sum(a,b);

printf("x=%dand y=%d",x,y);

void sum(x,y) {

int sum=x+y;

}

Here 5 and 2 are copied to the function and hence the local variables x and y store them but they

are not accessible from outside (main method).

int a=5;

change(a);

printf("%d",a); //gives output 5 instead of 0

void change(a){



138

a=0;

}

Now look here ‘a’ declared in function is different than declared in main method as we see that the

print of ‘a’ in main method even after function call shows that a=5 still, this shows that the address

of a, b in function is different than address of a, b in main method.

Now call by address or reference is just oppose of what we have gone through. In call by address,

the value is changed by any sort of changes either in function or in main method as in both cases

there is only one value with only one address for a variable.

A Simple program demonstrating structure: It takes data and print info of

whose address is “pokhara”.

A Simple program demonstrating structure:

It takes data and print info of whose address is “pokhara”

#include<stdio.h>

#include<conio.h>

#include<string.h>

struct employee {

char name[40];

char address[30];

int tel;

int salary;

struct yrofjoin {

int y;

} year;

}e1[100];

void main(){

int i;

int n;

printf("number of data");

scanf("%d",&n);

for(i=0;i<n;i++){

printf("Name\t");

scanf("%s",e1[i].name);

printf("Address\t");

scanf("%s",e1[i].address);

printf("Telephone\t");

scanf("%d",&e1[i].tel);

printf("Salary\t");

scanf("%d",&e1[i].salary);

printf("Year of Join\t");

scanf("%d",&e1[i].year.y);

}

for(i=0;i<n;i++){

if(strcmp(e1[i].address,"pokhara")==0) {



139

printf("\n the info of persons whose address is pokhara are");

printf("\n\t%s\t%s\t%d\t%d\t%d\n",e1[i].name,e1[i].address,e1[i].tel,e1[i].salary,e1[i].year.y);

}

}

getch();

}

FILE HANDLING

File pointer vs std

fscanf(stdin,"%d",&roll);

This is the next format of 'fscanf', here it will just take input from screen as stdin means standard

input i.e. input from screen but does nothing to file similar this 'stdout' by 'fprintf' prints data not

from file but from program to screen i.e. std means reading from input of user i.e. from screen to

program or printing to screen from program.

printf("Enter Phone number\n");

fscanf(stdin,"%s",&phone); // it takes input from screen to program

fprintf(fp1a," %s %s %d",name,add,phone); //this prints data from program to file

defined by ‘fp1a’

fscanf(fp1a," %s %s %d",name,add,phone); //it reads data from file defined by pointer

‘fp1a’

fprintf(stdout,"\n%s\t%s %d\n\n",name,add,phone); //it prints data from program to

screen

by above we come to two conclusions that if there is file pointer in the function that does connect

between program and file (it may be both reading and writing operations) however ‘std’ does

connection between program and screen, even though ‘fprintf’ or ‘fscanf’ seems to be only with file

handling. The other conclusion is that there must be certain relationship:

Screen Program File



140

From the above example, we see that at first phone number is taken from user as input to program

that is then stored in file and similarly, phone number is read from file to program then only printed

to screen, however direct printing and reading between screen and file is not possible.

PU2009 SPRING 6 B)

WAP to read name, author and price of 500 books to “library.dat” and then print info of books of

price > 300.

#include<stdio.h>

#include<conio.h>

struct library {

char book_name[35];

char book_author[15];

float book_price;

}element[500];

void main() {

int i;

FILE*fp;

printf("\nEnter Information of 500 Books\n");

fp=fopen("library.dat","w");//cause creating a new file or replacing if there is old file

for(i=0;i<500;i++) {

printf("Enter name:");

scanf("%s",&element[i].book_name);

printf("Enter Author:");

scanf("%s",&element[i].book_author);

printf("Enter Price:");

scanf("%f",&element[i].book_price);

fprintf(fp," %s %s %f",element[i].book_name,element[i].book_author,element[i].book_price);

}

fclose(fp);

fp=fopen("library.dat","r");

printf("\n Books whose price greater than Rs. 300\n");

printf("Name\t\t Author\t\t price\n");

for(i=0;i<500;i++) { // it is describe din B-part

// while(fp!=feof) { or while(fp!=NULL) { can be used instead this is described in A-part

fscanf(fp," %s %s %f",&element[i].book_name,&element[i].book_author,&element[i].book_price);

if(element[i].book_price>300) {

printf("%s\t%s\t\t0.3%f\n",element[i].book_name,element[i].book_author,element[i].book_price);

}

}

//TO HERE

fclose(fp);

getch();

}



141

Analysis:

Problem:

Here we need three variables to store name of book, author of book and price of book, but these

variables work only to a book so an array of 5800 book is required. We need to store all the

information of 500 books to a file and after that we have to read them and print the books having

price more than 300. So here an structure; a simple structure is necessary as there is no necessary of

branching structure elements as name of book, author and price are totally independent to each

other as well as they belong to different groups. The entity of that structure must be an array of 500

which can be used to read elements.

But for the particular case of above we use its advantage in printing too i.e. the 500 elements are

used for checking the conditio0n (price>300) and printing if satisfied.

B-part:

We know that there are 500 books so we directly did it, i.e. we continued our loop of i for 500 times

but in case we are unsure we follow A. In B, we are using the structure again from its object

elements initial value 0 to 499 to store the value from the file. If the condition is met, then it is

printed out. Instead of the array of structure’s object “element” we can use a single variable that

stores a value at a loop and check condition and work according to condition and for next loop the

variable stores next value and similarly so for goes on OR we may use array of variables without any

structure; it is because the array has noth8ing to do with the condition as each array check at each

time which is same to each variable at a time.

PU2009fall6 b)

Create a structure for the following data:

ID Name Address Salary Date of joining

dd mm yy

And write program to input 100 employees, save in emp.dat file and print info of employee who

have the address 'pokhara'.

#include<stdio.h>

#include<conio.h>

#include<string.h>

struct employee {

int id;

char name[40];

char address[30];

int salary;

struct yrofjoin {



142

int d;

int m;

int y;

} yoj;

}e1[100];

void main(){

int i;

FILE*fp;

fp=fopen("emp.dat","w");

for(i=0;i<100;i++){

printf("ID\t");

scanf("%d",&e1[i].id);

printf("Name\t");

scanf("%s",&e1[i].name);

printf("Address\t");

scanf("%s",&e1[i].address);

printf("Salary\t");

scanf("%d",&e1[i].salary);

printf("Day of Join\t");

scanf("%d",&e1[i].yoj.d);

printf("Month of Join\t");

scanf("%d",&e1[i].yoj.m);

printf("Year of Join\t");

scanf("%d",&e1[i].yoj.y);

fprintf(fp,"%d-%s-%s-%d-%d-%d-

%d/",e1[i].id,e1[i].name,e1[i].address,e1[i].salary,e1[i].yoj.d,e1[i].yoj.m,e1[i].yoj.y);

}

printf("\n\n");

fclose(fp);

fp=fopen("emp.dat","r");

printf("DATA STORED ABOUT EMPLOYEE WHO LIVES IN POKHARA\n");

printf("ID\t");

printf("Name\t\t");

printf("Address\t\t");

printf("Salary\t\t");

printf("Date of join\n\n");

for(i=0;i<100;i++){

fscanf(fp,"%d-%s-%s-%d-%d-%d-

%d/",&e1[i].id,&e1[i].name,&e1[i].address,&e1[i].salary,&e1[i].yoj.d,&e1[i].yoj.m,&e1[i].yoj.y);

if(strcmp(e1[i].address,"pokhara")==0) {

printf("%d\t%s\t%s\t%d\t%d-%d-

%d/",e1[i].id,e1[i].name,e1[i].address,e1[i].salary,e1[i].yoj.d,e1[i].yoj.m,e1[i].yoj.y);

}

}

fclose(fp);

getch();

}

We have used:-



143

fprintf(fp,"%d-%s-%s-%d-%d-%d-%d/",

e1[i].id,e1[i].name,e1[i].address,e1[i].salary,e1[i].yoj.d,e1[i].yoj.m,e1[i].yoj.y);

here

%d-%s-%s-%d-%d-%d-%d/

Is the format to store data, instead of – and / we can use other symbols too. Why are they used?

Let us suppose an input like

ID: 1

NAME: abc

ADDRESS: pokhara

SALARY: 500

DAY OF JOIN:1

MONTH OF JOIN:2

YEAR OF JOIN:2069

And if we put the format like this

%d%s%s%d%d%d%d

Without any space and symbols then in it will be save like this

1abcpokhara500122069

And while reading from file by

fscanf(fp,“%d%s%s%d%d%d%d”………………….);

%d will surely extract 1, but %s succeeding %d will extract abcpokhara because in the stored data

abcpokhara is a single string but here our intensition was to just extract NAME: abc, the second %s

succeeding it will extract no string as abc pokhara is already extracted by preceding %s and

500122069 is not a string.similarly the %d intentended to extract salary:500 will extract 500122069

and left %d%d%d for day, month,year will have nothing to extract.

So we need to use either appropriate symbols that has not been used in datas stored or we can

simply use single or double or more spaces or tabs according to complexity of data storage like this

%d %s %s %d %d %d %d

Here double space has been used, it is good not to use single space if we are storing string because

string like name can have space between first and last name.

Look carefully

%d-%s-%s-%d-%d-%d-%d/



144

We have used /, why think reasonably. It is because in file datas are added one-by-one in compact

manner i.e. if we store two datas like this:

ID: 1

NAME: abc

ADDRESS: pokhara

SALARY: 500

DAY OF JOIN:1

MONTH OF JOIN:2

YEAR OF JOIN:2069

ID: 2

NAME: xyz

ADDRESS: lekhnath

SALARY: 123

DAY OF JOIN:5

MONTH OF JOIN:11

YEAR OF JOIN:2066

And the format of saving in file is

“%d-%s-%s-%d-%d-%d-%d”

Then file it will be stored like this

1-abc-pokhara-500-1-2-20692-xyz-lekhnath-123-5-11-2066

Here you notice that in 20692, 2069 is year for first person and 2 is id of second person, but similar

to above, 20692 will be extracted as a single int type data and there will be no id to extract for

second person so we use

“ %d-%s-%s-%d-%d-%d-%d/”

which will change the above format of storing into

1-abc-pokhara-500-1-2-2069/2-xyz-lekhnath-123-5-11-2066

It is better if you use

“ /%d-%s-%s-%d-%d-%d-%d”

Here / is in front so that before each set of data is stored / is stored keeping a barrier between sets

of data, but use what you feel good and what program accepts without destroying the barrier

between set of data and also within the data of same set. We discussed about two barriers- one

internal between data of same set and another external between sets of data.

NOTE: In reading file by file_pointer, the last term of each installment of data in file must be

completely free from spaces or any symbols used as barrier like this

“ %d %s %f”

because file pointers read all (symbols too) showing that file is not feof i.e. it reads some symbols

like this

while(!feof(fp)){

fscanf(fp, “ %d %s %f ”);

fprintf(stdout, “ %d %s %f”);

}

Here even after all required values of data have been read and printed then ‘fscanf’ reads last

double space which shows ‘feof’ of ‘while’ is false (means ‘while-loop’ executes once more) even

after all required datas is already printed then ‘fprint’ is executed but the format does match as the



145

double spaces of file is unmatched to “ %d %s %f ” so the previous latest values of variables are

once more printed which means the last data of file is double printed. But next time, ‘feof’ is true

and ‘while-loop’ terminates.

Due fee

Prints student info whose due fee is less than 100

#include<stdio.h>

#include<conio.h>

void main(){

FILE *fp;

int n;

int roll;

float duefee;

int i;

char name[50];

fp=fopen("student.txt","w");

printf("How many input");

scanf("%d",&n);

for(i=0;i<n;i++) {

printf("Enter Student ID\t");

printf("Enter Student Name\t");

printf("Enter Duefee\n\n");


fscanf(stdin,"%s",&name);

fscanf(stdin,"%f",&duefee);

fprintf(fp," %d %s %f",roll,name,duefee);

}

fclose(fp);

fp=fopen("student.txt","r");

printf("STUDENTS WITH DUE FEES LESS THAN 100\n");

while(!feof(fp)){

fscanf(fp," %d %s %f",&roll,&name,&duefee);

if(duefee<100){

printf("ID\t");

printf("Name\t");

printf("Duefee\n\n");

fprintf(stdout,"\n%d\t%s\t%f\n",roll,name,duefee);

}

}

fclose(fp);

getch();

}



146

PU2002 Consumed charged data merged with file handling: WAP that stores following data

of customer of electricity authority in file ‘elect.dat’: id, name, old reading, new reading and amount

to be paid.

ELECRTIC DATA MERGED WITH FILE HANDLING

WAP that stores following data of customer of electricity authority in file ‘elect.dat’:

Id Name Old reading New reading Amount to be

paid

Hints: The amount to be paid is calculated by calculating the units consumed by subtracting ‘New

reading’ to ‘Old reading’ and following given list shows the range of units and rate as:

Range of units Rate per unit

0-200 4.5

201-400 5.5

400-above 7.5

Use this file to store data from user until user supplies ‘0’ as Id then after display the above list.

#include<stdio.h>

#include<conio.h>

struct customer{

int id;

char name[40];

int old_reading;

int new_reading;

float amount;

}cus[1000];

float calculation(int x){

float amount;

if(x<=200) {

amount=x*4.5;



147

}

else if(x>200 && x<=400) {

amount=200*4.5;

x=x-200;

amount=amount+(x*5.5);

}

else {

amount=200*4.5+200*5.5;

x=x-400;

amount=amount+x*7.5;

}

return (amount);

}

void main(){

int i=-1,value;

FILE*fp;

fp=fopen("elect.dat","w");

do {

i++;

printf("CUSTOMER ID\t");

scanf("%d",&cus[i].id);

printf("Name\t");

scanf("%s",&cus[i].name);

printf("Old Reading\t");

scanf("%d",&cus[i].old_reading);

printf("New Reading\t");

scanf("%d",&cus[i].new_reading);

value=cus[i].new_reading-cus[i].old_reading;

cus[i].amount=calculation(value);

fprintf(fp," %d %s %d %d

%.2f",cus[i].id,cus[i].name,cus[i].old_reading,cus[i].new_reading,cus[i].amount);/* we are using

double space as the string has scope over symbols like / or-*/

}

while((cus[i].id)!=0);

printf("\n\n");

fclose(fp);

fp=fopen("elect.dat","r");

printf("DATA STORED ABOUT CUSTOMERS OF ELECTRICITY AUTHORITY\n");

printf("ID\t");

printf("Name\t\t");

printf("Old reading\t\t");

printf("New reading\t\t");

printf("Amount to be paid\n\n");

i=0; /*previously 'i' was used to store to file now for reading from file so must be initialized as zero

or any value or nothing too works, even other variables can be declared but similar change to all

preceding program*/

while(!feof(fp)) {

fscanf(fp," %d %s %d %d

%f",&cus[i].id,&cus[i].name,&cus[i].old_reading,&cus[i].new_reading,&cus[i].amount);

printf("\n");



148

printf("%d\t%s\t%d\t%d\t%.2f",cus[i].id,cus[i].name,cus[i].old_reading,cus[i].new_reading,cus[i].am

ount);

}

fclose(fp);

getch();

}

Here ‘calculation()’ function returns amount to be paid in float type whereas it takes consumed units

as input. For calculation, here we the very simple process among the ways we discussed in

ELECRTICITY AUTHORITY sector.

As we see that do-while loop has condition to be checked which contains ‘i’ so here the value of ‘i’

must not increase or simply updated just before while-condition as this directly hampers the

condition of while so we have incremented ‘i’ in line following do-loop, but as we know ‘i’ must

starts from 0 and there is increment just after do-loop, so initially ‘i’ is assigned -1. The while-loop

only continues if the Id provided by user previously is not zero then continues with new input from

user saves it in file and then only checks if the id sent by user is zero or not, i.e. in file, we find that

the last stored data must have zero as id. Besides, it includes all data like name, readings, amount for

the customer to which our intension was not to be saved in file. This is the defect of the above

program.

This can be removed by adding a ‘if=statement’ checking whether the input Id is zero or not, if zero

then goto outside else continue by using goto inside:, like below:

#include<stdio.h>

#include<conio.h>

struct customer{

int id;

char name[40];

int old_reading;

int new_reading;

float amount;



149

}cus[100];

float calculation(int x){

float amount;

if(x<=200) {

amount=x*4.5;

}

else if(x>200 && x<=400) {

amount=200*4.5;

x=x-200;

amount=amount+(x*5.5);

}

else {

amount=200*4.5+200*5.5;

x=x-400;

amount=amount+x*7.5;

}

return (amount);

}

void main(){

int i=0,net_unit=0;

FILE*fp;

fp=fopen("elect.dat","w");

inside:

printf("CUSTOMER ID\t");

scanf("%d",&cus[i].id);

if((cus[i].id)==0) goto outside;

printf("Name\t");

scanf("%s",&cus[i].name);

printf("Old Reading\t");

scanf("%d",&cus[i].old_reading);

printf("New Reading\t");

scanf("%d",&cus[i].new_reading);

net_unit=cus[i].new_reading-cus[i].old_reading;

cus[i].amount=calculation(net_unit);

fprintf(fp," %d %s %d %d

%.2f",cus[i].id,cus[i].name,cus[i].old_reading,cus[i].new_reading,cus[i].amount);

goto inside;

outside:

printf("\n\n");

fclose(fp);

fp=fopen("elect.dat","r");

printf("DATA STORED ABOUT CUSTOMERS OF ELECTRICITY AUTHORITY\n");

printf("ID\t");

printf("Name\t\t");

printf("Old reading\t\t");

printf("New reading\t\t");

printf("Amount to be paid\n\n");

i=0; //here it is noted that i=0 for each time the loop of while so a static counter variable can be

usrd to store various data at varous steps



150

while(!feof(fp)) {

fscanf(fp," %d %s %d %d

%f",&cus[i].id,&cus[i].name,&cus[i].old_reading,&cus[i].new_reading,&cus[i].amount);

printf("%d\t%s\t%d\t%d\t%.2f\n",cus[i].id,cus[i].name,cus[i].old_reading,cus[i].new_reading,cus[i].

amount);

}

fclose(fp);

getch();

}

Study in above program; some changes taking place to value initialization of ‘i’ carefully.

PROJECTS

BOOKS

#include<stdio.h>

#include<conio.h>

void main(){

FILE *fp;

int n;

int code;

float price;

int i;

char name[50];

char book[50];

fp=fopen("booksinfo.txt","a");

printf("How many inputs?\t");

scanf("%d",&n);

for(i=0;i<n;i++) {

printf("Enter Book Code\n");



151

printf("Enter Author Name\n");

printf("Enter Book Name\n");

printf("Enter Cost Of Book\n");

fscanf(stdin,"%d",&code);


fscanf(stdin,"%s",&book);

fscanf(stdin,"%f",&price);

fprintf(fp," %d %s %s %f",code,name,book,price);

}

printf("\n\n");

fclose(fp);

fp=fopen("booksinfo.txt","r");

printf("DATA STORED ABOUT BOOKS\n");

printf("CODE\t");

printf("NAME\t\t");

printf("BOOK\t\t");

printf("PRICE\n\n");

while(!feof(fp)){

fscanf(fp," %d %s %s %f",&code,&name,&book,&price);

fprintf(stdout,"\n%d\t%s\t%s\t%f\n",code,name,book,price);

}

fclose(fp);

getch();

}



152

PHONE BOOK

Store in a file “phonebook.txt” following:

Name Address Telephone

Using ‘switch-case’ statement, keep insert option and search option that searches info of the file in

regard to the number or string or even a character given by user. Here it means that few characters

also does working as an input for searching.

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

int main() {

long double phone;/* these numbers have high range

unsigned long int range from 0 to 4294967295 i.e. 10 digits but lesser than mobile numbers

even float is small

so long double of 80 bits ranging 3.4e-4932 to 1.1e+4932

is used */

char phone_check[25]; // while comparing it is possible with string or character only

int num,found;

FILE *fp1a;

FILE*fp2a;

FILE*fp3a;

FILE*fp4a;

FILE*fp5a;

char name[25],add[50],name_check[25],add_check[50],phone_str[30];

printf("If you are using first time then goto option 1\n\n");

head:

found=0;// for each time head is referred a new initialization is required


printf("[1]\tto add info\n");

printf("[2]\tto look info list\n");


printf("[4]\tto search by address\n");

printf("[5]\tto search by telephone\n");


printf("-------------------------------------------\n");

scanf("%d",&num);

printf("---------------------------------------------------------\n");

switch(num) {

case 1:




153

printf("Enter Address\n");

printf("Enter Phone number\n");


fscanf(stdin,"%s",&add);

fscanf(stdin,"%lf",&phone);

fp1a=fopen("phonebook.txt","a");

fprintf(fp1a," %s %s %.0lf",name,add,phone); //".0" as phone number has no value after

decimal

fclose(fp1a);

if(getch())goto head;

case 2:

printf("INFO OF FRIENDS YOU KNOW\n");

printf("\n\n--------------------------------------------------------------\n");

printf("NAME ");

printf("ADDRESS ");

printf("PHONE NUMBER\t\n");

fp2a=fopen("phonebook.txt","r");

while(!feof(fp2a)){

fscanf(fp2a," %s %s %lf",&name,&add,&phone);

fprintf(stdout,"\n%s\t%s\t%.0lf\n\n",name,add,phone);

}

fclose(fp2a);


case 3:

printf("Enter name to be searched\t");

scanf("%s",&name_check);


printf("\n\n\n\n INFO OF FRIENDS beginning with name \'%s\'\t\n",name_check);

printf("\n\n--------------------------------------------------------------\n");

printf("NAME ");

printf("ADDRESS ");

printf("PHONE\t");

while(!feof(fp3a)){


if(strncmp(name,name_check,strlen(name_check))==0) {




154

found++;

}

}

fclose(fp3a);

if(found==0) {

printf("\nNOT FOUND\n");

}


case 4:

printf("Enter address to be searched\t");

scanf("%s",&add_check);


printf("\n\n\n\n INFO OF FRIENDS beginning with address \'%s\'\t\n",add_check);

printf("\n\n--------------------------------------------------------------\n");

printf("NAME ");

printf("ADDRESS ");

printf("PHONE\t");

while(!feof(fp4a)){


if(strncmp(add,add_check,strlen(add_check))==0) {


found++;

}

}

fclose(fp4a);

if(found==0) {


}


case 5:

printf("Enter phone to be searched\t");

scanf("%s",&phone_check);


printf("\n\n\n\n INFO OF FREIENDS beginning with phone \'%s\'\t\n",phone_check);



155

printf("\n\n--------------------------------------------------------------\n");

printf("NAME ");

printf("ADDRESS ");

printf("PHONE\t");

while(!feof(fp5a)){

fscanf(fp5a," %s %s %s",&name,&add,&phone_str); //here '%s' at last of format works

because numbers can be string but string cannot be numbers

if(strncmp(phone_str,phone_check,strlen(phone_check))==0) {

fprintf(stdout,"\n%s\t%s\t%s\n\n",name,add,phone_str);

found++;

}

}

fclose(fp5a);

if(found==0) {


}


case 6:

printf("THANKS TO YOU\n GOOD BYE !");

if(getch()){

exit(0);

}

}

return 1;

}



156



157



158

MARKSHEET BY HARDER

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

FILE *fp;

FILE *fp1;

FILE*fp2;

int num;

int number,a;

int roll;

int roll_array[1000];

int j=0,k=0;

int i=0;

int roll_t;float mark1_t;float mark2_t;float mark3_t;float mark4_t;float mark5_t;float perct_t;

int rol;

int abc=1;

float mark1,mark2,mark3,mark4,mark5,percent;

float

mark1_array[100],mark2_array[100],mark3_array[100],mark4_array[100],mark5_array[100],perct_

array[100];

float mar1,mar2,mar3,mar4,mar5,perc;

char name[100];

char nam[30];

char name_array[1000][30];

char dummy[20];

char nam[30];

int main() {

printf("If it is your first time then goto option 1\n\n");

head:





printf("-------------------------------------------\n");

scanf("%d",&num);

printf("-------------------------------------------\n");

switch(num) {

case 1:




159


fp=fopen("marksheet.txt","a");

for(a=0;a<number;a++) {





printf("Enter Marks in Electrical\n");

printf("Enter Marks in Thermal\n");

printf("Enter Marks in Physics\n");



fscanf(stdin,"%f",&mark1);





percent=(mark1+mark2+mark3+mark4+mark5)/5;

fprintf(fp," %d %s %f %f %f %f %f

%f",roll,name,mark1,mark2,mark3,mark4,mark5,percent);

}

fclose(fp);

if(getch()) goto head;

case 2:

fp=fopen("marksheet.txt","r");

while(!feof(fp)){

fscanf(fp," %d %s %f %f %f %f %f

%f",&roll_array[i],&name_array[i],&mark1_array[i],&mark2_array[i],&mark3_array[i],&mark4_arra

y[i],&mark5_array[i],&perct_array[i]);

i++;

}

fclose(fp);

//Sorting begins-descending by bubble method

for(j=1;j<i;j++) {

for(k=1;k<=i-j;k++) {

if(perct_array[k-1]<perct_array[k]) {

roll_t=roll_array[k-1];

roll_array[k-1]=roll_array[k];

roll_array[k]=roll_t;

strcpy(dummy,name_array[k-1]);

strcpy(name_array[k-1],name_array[k]);

strcpy(name_array[k],dummy);

mark1_t=mark1_array[k-1];

mark1_array[k-1]=mark1_array[k];

mark1_array[k]=mark1_t;




160












perct_t=perct_array[k-1];

perct_array[k-1]=perct_array[k];

perct_array[k]=perct_t;

}

}

}

// Rank List

fp1=fopen("ascending list of mark_sheet.txt","w");

for(j=0;j<i;j++) {

fprintf(fp1," %d %s %f %f %f %f %f

%f",roll_array[j],name_array[j],mark1_array[j],mark2_array[j],mark3_array[j],mark4_array[j],mark5

_array[j],perct_array[j]);

}

fclose(fp1);


printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");

printf("Electrical ");

printf("Thermal ");

printf("Physics ");

printf("PERCENT");

fp2=fopen("ascending list of mark_sheet.txt","r");

while(!feof(fp)){

fscanf(fp," %d %s %f %f %f %f %f %f",&rol,&nam,&mar1,&mar2,&mar3,&mar4,&mar5,&perc);

fprintf(stdout,"\n%d\t%d %s\t %.2f\t %.2f %.2f %.2f \t%.2f

%.2f\n\n\n",abc,rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

abc++;

}

if(getch()) goto head;



161

case 3:

printf("Thanks for your attention , please also refer 'by easier' and 'updated' method");


}

getch();

return 1;

}



162

MARKSHEET BY EASIER

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

FILE*fp;

FILE*fp1;

FILE *fp3;

FILE*fp4;

int rola;

int num;

int rol;

int check;

int rol3;

int number,a;

int roll;

int abca;

int founda;

int count;

int abc3;

int found3;

int abc;

float mara1,mara2,mara3,mara4,mara5,perca;

float mark1,mark2,mark3,mark4,mark5,percent;

float mar31,mar32,mar33,mar34,mar35,perc3;

float mar1,mar2,mar3,mar4,mar5,perc;

char nama[30];

char name[100];

char c;

char nam3[30];



163

char nam[30];

char checkk[30];

char file_status[10]="old";

int main() {


if(fp==NULL) { printf("No file was created till now\nNow created...\n\n");

fclose(fp);

fp=fopen("marksheet.txt","w");

fclose(fp);

strcpy(file_status,"new");

}

A:







abca=0;

founda=0;

abc3=0;

found3=0;

abc=1;

printf("-------------------------------------------\n");

scanf("%d",&num);

printf("-------------------------------------------\n");

switch(num) {

case 1:




count=0; //for each new entered number or each new loop of 'i' count=0









164











fp1=fopen("temp of mark_sheet.txt","w");

if(strcmp(file_status,"old")==0){

while(!feof(fp)){

fscanf(fp," %d %s %f %f %f %f

%f %f",&rol,&nam,&mar1,&mar2,&mar3,&mar4,&mar5,&perc);

if(count==0){

if(percent<perc) {


%f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

}

else {

count++;


%f",roll,name,mark1,mark2,mark3,mark4,mark5,percent); //this means that new data is entered

fprintf(fp1," %d %s %f %f %f %f %f %f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

}

}

else {



}

} //while-loop brace

if(count==0) fprintf(fp1," %d %s %f %f %f %f %f

%f",roll,name,mark1,mark2,mark3,mark4,mark5,percent); /*this means that new data is entered

for the case comparison is done to all values of files but all data of files if found descending than

this new value thewn at last new value is added i.e. in this case after while loop terminates*/

}



165

else { /* if file is new then just print it the value as fp in reading newly created file gives some

unnecessary old values so it is done*/

fprintf(fp1," %d %s %f %f %f %f %f %f",roll,name,mark1,mark2,mark3,mark4,mark5,percent);

//this means that new data is entered

strcpy(file_status,"old");

}

fclose(fp);

fclose(fp1);


fp1=fopen("temp of mark_sheet.txt","r");

while((c=getc(fp1))!=EOF) {

putc(c,fp);

}

fclose(fp);

fclose(fp1);

}

if(getch())goto A;

case 2:


printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");


while(!feof(fp)){

fscanf(fp," %d %s %f %f %f %f %f %f",&rol,&nam,&mar1,&mar2,&mar3,&mar4,&mar5,&perc);



abc++;

}

fclose(fp);

if(getch())goto A;

case 3:


scanf("%d",&check);



166

fp3=fopen("marksheet.txt","r");

printf("\n\n\n\n MARKSHEET OF STUDENTS with roll \'%d\'\t\n",check);

printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");

while(!feof(fp3)){

abc3++;

fscanf(fp3," %d %s %f %f %f %f %f

%f",&rol3,&nam3,&mar31,&mar32,&mar33,&mar34,&mar35,&perc3);

if(rol3==check) {


%.2f\n\n\n",abc3,rol3,nam3,mar31,mar32,mar33,mar34,mar35,perc3);

found3++;

}

}

fclose(fp3);

if(found3==0) {


}

if(getch())goto A;

case 4:


scanf("%s",&checkk);

fp4=fopen("marksheet.txt","r");

printf("\n\n\n\n MARKSHEET OF STUDENTS beginning with name \'%s\'\t(RANK

WISE)\n",checkk);

printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");

while(!feof(fp4)){

abca++;

fscanf(fp4," %d %s %f %f %f %f %f

%f",&rola,&nama,&mara1,&mara2,&mara3,&mara4,&mara5,&perca);



167

if(strncmp(nama,checkk,(strlen(checkk)))==0) {


%.2f\n\n\n",abca,rola,nama,mara1,mara2,mara3,mara4,mara5,perca);

founda++;

}

}

fclose(fp4);

if(founda==0) {


}

if(getch())goto A;

case 5:

printf("THANKS TO YOU FOR VISITING THIS MARKSHEET PROGRAM");

if(getch()){

exit(0);

}

}

getch();

return 1;

}



168



169

UPDATED MARKSHEET

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

FILE *fp;

FILE *fp1a;

FILE*fp1b;

FILE*fp1c;

FILE *fp2a;

FILE*fp1d;

FILE *fp3a;

FILE *fp4a;

FILE *fp5a;

FILE *fp5b;

FILE *fp5c;

FILE *fp5d;

FILE *fp6a;



170

FILE *fp6b;

FILE *fp6c;

FILE *fp6d;

int num,number,a,roll,rol,roll_check,roll_delete;

float mar1,mar2,mar3,mar4,mar5,perc,mark1,mark2,mark3,mark4,mark5,percent;

char c,nam[30],name[100],name_check[30],name_delete[15],file_status[5]="old";

int main() {


if(fp==NULL) { printf("No file was created till now\nNow created...\n\n");

fclose(fp);


fclose(fp);


}

printf("If your are using first time then goto option 1\n\n");

head:






printf("[5]\tto delete by roll\n");

printf("[6]\tto delete by name\n");


printf("-------------------------------------------\n");

int count,found=0,abc=1;

scanf("%d",&num);

printf("---------------------------------------------------------\n");

switch(num) {

case 1:




count=0; //same to previous 'easier method' refer it











171








fp1a=fopen("marksheet.txt","r");

fp1b=fopen("temp of mark_sheet.txt","w");


while(!feof(fp1a)){

fscanf(fp1a," %d %s %f %f %f %f %f

%f",&rol,&nam,&mar1,&mar2,&mar3,&mar4,&mar5,&perc);

if(count==0){

if(percent<perc) {

fprintf(fp1b," %d %s %f %f %f %f %f


}

else {

count++;


%f",roll,name,mark1,mark2,mark3,mark4,mark5,percent);

fprintf(fp1b," %d

%s %f %f %f %f %f %f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

}

}

else fprintf(fp1b," %d %s %f %f

%f %f %f %f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

} //while-loop terminal brace

if(count==0) fprintf(fp1b," %d %s %f %f %f %f

%f %f",roll,name,mark1,mark2,mark3,mark4,mark5,percent); /* same to old method: this means

that new data is entered for the case comparison is done to all values of files but all data of files if

found descending than this new value thewn at last new value is added i.e. in this case after while

loop terminates*/

}

else { /* if file is new then just print it the value as fp in reading

newly created file gives some unnecessary old values so it is done*/


%f",roll,name,mark1,mark2,mark3,mark4,mark5,percent); //this means that new data is entered


}

fclose(fp1a);

fclose(fp1b);

fp1c=fopen("marksheet.txt","w");



172

fp1d=fopen("temp of mark_sheet.txt","r");

while((c=getc(fp1d))!=EOF) {

putc(c,fp1c);

}

fclose(fp1c);

fclose(fp1d);

} // terminal of for-loop


case 2:


printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");


while(!feof(fp2a)){





}

fclose(fp2a);


case 3:


scanf("%d",&roll_check);

printf("\n\n\n\n MARKSHEET OF STUDENTS beginning with \'%d\'\t(RANK

WISE)\n",roll_check);

printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");




173

while(!feof(fp3a)){

count++;



if(rol==roll_check) {


%.2f\n\n\n",count,rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

found++;

}

}

fclose(fp3a);

if(found==0) {


}


case 4:


scanf("%s",&name_check);


printf("\n\n\n\n MARKSHEET OF STUDENTS beginning with \'%s\'\t(RANK

WISE)\n",name_check);

printf("\n\n--------------------------------------------------------------\n");

printf("RANK ");

printf("ROLL ");

printf("NAME\t");


printf("Maths ");


printf("Thermal ");

printf("Physics ");

printf("PERCENT");

while(!feof(fp4a)){

count++;



if(strncmp(nam,name_check,(strlen(name_check)))==0) {


%.2f\n\n\n",count,rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

found++;

}

}

fclose(fp4a);

if(found==0) {


}



174


case 5:

printf("Enter Roll number whose marksheet is to be deleted");

scanf("%d",&roll_delete);



while(!feof(fp5a)){



if(roll_delete==rol){

printf("Marksheet is updated\n");

}

else {

fprintf(fp5b," %d %s %f %f %f %f %f %f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

}

}

fclose(fp5a);

fclose(fp5b);




putc(c,fp5c);

}

fclose(fp5c);

fclose(fp5d);

if(getch()){ goto head;

}

case 6:

printf("Enter name beginning with to be deleted");

scanf("%s",name_delete);



while(!feof(fp6a)){



if(strncmp(nam,name_delete,strlen(name_delete))==0){

printf("Marksheet is updated\n");

}

else {

fprintf(fp6b," %d %s %f %f %f %f %f %f",rol,nam,mar1,mar2,mar3,mar4,mar5,perc);

}

}

fclose(fp6a);

fclose(fp6b);



175




putc(c,fp6c);

}

fclose(fp6c);

fclose(fp6d);

if(getch()){ goto head;

}

case 7:

printf("THANKS TO YOU\n GOOD BYE !");

if(getch()){

exit(0);

}

}

return 1;

}



176



177



178

STORES

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

FILE *fp;

FILE*fp1;

FILE*fp2;

FILE*fp3;

FILE*fp11;

struct tempcalculation {

char tempname[40];

int tempquantity;

float temprate;

}temp[100];

int num;

int code;

int stock;

int costcode;

int costquantity;

int new;

int n,i;

int code;

int na;

int naa;

int neew;

int tempi;

float price;

float rate;



179

float total;

char item[100];

char name[40];

char item[100];


int main(){

fp=fopen("stores.txt","r");

if(fp==NULL) { printf("No stores file was created till now\nNow created...\n\n");

fclose(fp);

fp=fopen("stores.txt","w");

fclose(fp);


}

printf("****************************************WELCOME TO

STORES*****************************\n\n\n");

printf("-----------------------------------------------------------------------------------\n");

printf("Enter your full name:-\t");

scanf("%[^\n]",name);

main:

printf("\n\n----------------------------------\n");


printf("[1]\tto look list\n");

printf("[2]\tto calculate\n");

printf("[3]\tto add items\n");

printf("[4]\tto change stock\n");

printf("[5]\tto add stock\n");

printf("[6]\tto exit\n\n");

printf("---------------------------------------------\n");

scanf("%d",&num);

printf("---------------------------------------\n");

tempi=0;//this must be initialized for each new install of calculation

switch (num) {

case 1:


printf("DATA STORED ABOUT ITEMS\n\n");

printf("\n\n-----------------------------------------------------------------------------------\n");

printf("CODE\t");

printf("ITEM NAME\t\t");

printf("PRICE\t\t");

printf("STOCK\n\n");

while(!feof(fp)){

fscanf(fp," %d %s %f %d",&code,&item,&price,&stock);

fprintf(stdout,"\n%d\t%s\t\t%f\t\t%d\n",code,item,price,stock);

}

fclose(fp);

if(getch()){



180

goto main;

}

case 2:

printf("CALCULATION\n");

tempcheck:

printf("Enter[0/0] to end\n");

printf("Enter [code]/[quantity]\t\t");

scanf("%d/%d",&costcode,&costquantity);

if(costcode==0 && costquantity==0 ) {

total=0;

printf("YOUR COST\n\n");

printf("-----------------------------------------------------------------------------\n");

printf("ITEMS\t\t\tRATE\t\t\tQUANTITY\t\tSUM\n");

printf("----------------------------------------------------------------------------\n");

for(i=0;i<tempi;i++) {

printf("%s\t\t",temp[i].tempname);

printf("%f\t\t",temp[i].temprate);

printf("%d\t\t",temp[i].tempquantity);

printf("%f\n",temp[i].temprate*temp[i].tempquantity);

total=total+(temp[i].temprate*temp[i].tempquantity);

}

printf("--------------------------------------------------\n");

printf("\t\t\t\t\t\tTOTAL==Rs.%.2f\n",total);

printf("Checked by :\t%s",name);

if(getch()){

goto main;

}

}

else {


fp1=fopen("tempofstore.txt","w");

while(!feof(fp)) {


if(costcode==code){

if(costquantity>stock){

printf("\n%s is not in stock\n",item);

fprintf(fp1," %d %s %f %d",code,item,price,stock);

continue;

}

new=stock-costquantity;

fprintf(fp1," %d %s %f %d",code,item,price,new);

strcpy(temp[tempi].tempname,item);//temp[tempi].tempname=item;

temp[tempi].temprate=price;

temp[tempi].tempquantity=costquantity;

tempi++;

}



181

else {


}

}

fclose(fp);

fclose(fp1);

fp3=fopen("tempofstore.txt","r");

fp2=fopen("stores.txt","w");

while(!feof(fp3)){

fscanf(fp3," %d %s %f %d",&code,&item,&price,&stock);


}

fclose(fp3);

fclose(fp2);

goto tempcheck;

}

case 3:

fp=fopen("stores.txt","a");


scanf("%d",&n);

for(i=0;i<n;i++) {

printf("Enter Code\n");

printf("Enter Item Name\n");

printf("Enter Rate\n");

printf("Enter Stocks\n");


fscanf(stdin,"%s",&item);

fscanf(stdin,"%f",&rate);

fscanf(stdin,"%d",&stock);

fprintf(fp," %d %s %f %d",code,item,rate,stock);

}

fclose(fp);

if(getch()){

goto main;

}

case 4:

printf("Enter code\t");

scanf("%d",&na);

printf("Change it by\t");



182

scanf("%d",&new);



while(!feof(fp)){


if(na==code){

fprintf(fp1," %d %s %f %d",code,item,price,new);

}

else


}

fclose(fp);

fclose(fp1);



while(!feof(fp1)){


fprintf(fp," %d %s %f %d",code,item,price,stock);

}

fclose(fp);

fclose(fp1);

if(getch()){

goto main;

}

case 5:


scanf("%d",&naa);

printf("Add it by\t");

scanf("%d",&neew);



while(!feof(fp)){


if(naa==code){

neew=stock+neew;

fprintf(fp11," %d %s %f %d",code,item,price,neew);

}

else


}

fclose(fp);

fclose(fp11);




183


while(!feof(fp1)){


fprintf(fp," %d %s %f %d",code,item,price,stock);

}

fclose(fp);

fclose(fp1);

if(getch()){

goto main;

}

case 6:

printf("\n\n------------------------------------------------------\n");

printf("THANKS %s FOR YOUR SERVICE",name);

printf("\n\n----------------------------------------\n");

if(getch()){

exit(1); } }return 1;}



184



185

UPDATED STORES

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

FILE *fp;

FILE *fp1a;

FILE *fp2a;

FILE*fp2b;

FILE*fp2c;

FILE*fp2d;

FILE*fp3aa;

FILE*fp3a;

FILE*fp3b;

FILE*fp3c;

FILE*fp3d;

FILE*fp4a;

FILE*fp4b;

FILE*fp4c;

FILE*fp4d;

FILE*fp5a;

FILE*fp5b;

FILE*fp5c;

FILE*fp5d;

FILE*fp6a;

FILE*fp6b;

FILE*fp6c;

FILE*fp6d;

FILE*fp7a;

FILE*fp8a;

struct tempcalculation {

char tempname[40];

int tempquantity;

float temprate;

}temp[100];

int num;

int code;

int stock;

int costcode;

int costquantity;

int new;

int n,i;

int code;

int na;

int naa;

int neew;

int tempi;

int new_stock,count=0,codef,stockf,del,na,naa,neew,code_search;



186

float price;

float rate;

float total;

float ratef;

float vat;

char item[100];

char name[40];

char itemf[100],c,item_search[15];


int main(){


if(fp==NULL) { printf("No stores file was created till now\nNow created...\n\n");

fclose(fp);


fclose(fp);


}

printf("***********************************WELCOME TO

STORES*****************************\n\n\n");

printf("**********************Programmer-Devendra Man

Palikhe**************************\n\n");

printf("------------------------------------------------------------------------------\n");

printf("Enter your full name:-\t");


main:

printf("\n\n----------------------------------------\n");


printf("[1]\tto look list\n");

printf("[2]\tto calculate\n");

printf("[3]\tto add items\n");

printf("[4]\tto delete item\n");

printf("[5]\tto change stock\n");

printf("[6]\tto add stock\n");

printf("[7]\tto search item by code\n");

printf("[8]\tto search item by item_name\n");

printf("[9]\tto exit\n\n");

printf("-------------------------------------------\n");

scanf("%d",&num);

printf("-------------------------------------------\n");

tempi=0;//this must be initialized for each new install of calculation

vat=0;

switch (num) {

case 1:

printf("\n");



187

fp1a=fopen("stores.txt","r");

printf("DATA STORED ABOUT ITEMS\n\n");

printf("\n\n--------------------------------------------------------------\n");

printf("CODE\t");




while(!feof(fp1a)){

fscanf(fp1a," %d %s %f %d",&code,&item,&price,&stock);


}

fclose(fp1a);

if(getch()){

goto main;

}

case 2:

printf("CALCULATION\n");

tempcheck:

printf("Enter[0/0] to end\n");

printf("Enter [code]/[quantity]\t\t");

scanf("%d/%d",&costcode,&costquantity);

if(costcode==0 && costquantity==0 ) {

total=0;

printf("YOUR COST\n\n");

printf("--------------------------------------------------\n");

printf("ITEMS\t\t\tRATE\t\t\tQUANTITY\tSUM\n");

printf("--------------------------------------------------\n");

for(i=0;i<tempi;i++) {

printf("%s\t\t",temp[i].tempname);

printf("%.2f\t\t",temp[i].temprate);

printf("%d\n",temp[i].tempquantity);

printf("%.2f\n",temp[i].temprate*temp[i].tempquantity);

total=total+(temp[i].temprate*temp[i].tempquantity);

}

printf("--------------------------------------------------\n");

printf("If any, then include VAT:\t");

scanf("%f",&vat);

printf("\t\t\t\t\t\tNET TOTAL==Rs.%.2f\n",total+(total*vat/100));

printf("\t\t\t\t\t\tVAT==Rs.%.2f\n",total*vat/100);

printf("\t\t\t\t\t\tTOTAL==Rs.%.2f\n",total+(total*vat/100));

printf("Checked by :\t%s",name);

if(getch()){

goto main;

}

}

else {




188

fp2b=fopen("temp of stores.txt","w");

while(!feof(fp2a)){


if(costcode==code){

if(costquantity>stock){

printf("\n%s is not in stock\n",item);

fprintf(fp2b," %d %s %f %d",code,item,price,stock);

continue;

}

new_stock=stock-costquantity;

fprintf(fp2b," %d %s %f %d",code,item,price,new_stock);

strcpy(temp[tempi].tempname,item);//temp[tempi].tempname=item;

temp[tempi].temprate=price;

temp[tempi].tempquantity=costquantity;

tempi++;

}

else {


}

}

fclose(fp2a);

fclose(fp2b);

fp2c=fopen("temp of stores.txt","r");

fp2d=fopen("stores.txt","w");

while(!feof(fp2c)){

fscanf(fp2c," %d %s %f %d",&code,&item,&price,&stock);

fprintf(fp2d," %d %s %f %d",code,item,price,stock);

}

fclose(fp2c);

fclose(fp2d);

goto tempcheck;

}

case 3:


scanf("%d",&n);

for(i=0;i<n;i++) {

next:

count=0; //for each new entered input or each new loop of 'i' count=0

printf("Enter Code\t");

printf("Enter Item Name\t");

printf("Enter Rate\t");

printf("Enter Stocks\n\n");


fscanf(stdin,"%s",&item);

fscanf(stdin,"%f",&rate);

fscanf(stdin,"%d",&stock);

fp3aa=fopen("stores.txt","r");

while(!feof(fp3aa)){



189

fscanf(fp3aa," %d %s %f %d",&codef,&itemf,&ratef,&stockf);

if(code==codef) {

printf("%d code is already in used, so it is invalid and not responded\n",code);

i++;

goto next;

}

}

fclose(fp3aa);

fp3a=fopen("temp of stores.txt","w");

fp3b=fopen("stores.txt","r");


while(!feof(fp3b)){

fscanf(fp3b," %d %s %f %d",&codef,&itemf,&ratef,&stockf);

if(count==0){

if(code<codef) { /*previously in markshhet data of file were only printed here both

because here it is ascending purpose but in marksheet just opposite or to maintain static we can

change '<' to '>'*/

fprintf(fp3a," %d %s %f %d",code,item,rate,stock);

fprintf(fp3a," %d %s %f %d",codef,itemf,ratef,stockf);/*we can directly print the

data from file too as if in a assecnding list a new number is added then the position of previous

value replaced by new number changes by +1 and similarly for others but here to others we do by

illogical way that is one by one*/

count++;

}

else {/*it is also reverse to marksehet*/

fprintf(fp3a," %d %s %f %d",codef,itemf,ratef,stockf);

}

}

else fprintf(fp3a," %d %s %f %d",codef,itemf,ratef,stockf);/*when the new data is

already insetred given by count=1 then just copying data from file to another file*/

} //while loop's terminal

if(count==0) fprintf(fp3a," %d %s %f %d",code,item,rate,stock); /*this statement is also

found in marksheet both work for same purpose; the statement executes when the new data has

no position in between or infront of data of file; here for ascending case and in marksheet for

descending*/

}

else {

fprintf(fp3a," %d %s %f %d",code,item,rate,stock);


}

fclose(fp3a);

fclose(fp3b);

fp3c=fopen("stores.txt","w");

fp3d=fopen("temp of stores.txt","r");


putc(c,fp3c);

}

fclose(fp3c);

fclose(fp3d);

}



190

if(getch()){

goto main;

}

case 4:

printf("enter code of item to be deleted\t");

scanf("%d",&del);



while(!feof(fp4a)){

fscanf(fp4a," %d %s %f %d",&codef,&itemf,&ratef,&stockf);

if(del==codef){ }

else

{

fprintf(fp4b," %d %s %f %d",codef,itemf,ratef,stockf);

}

}

fclose(fp4b);

fclose(fp4a);

fp4c=fopen("stores.txt","w");

fp4d=fopen("temp of stores.txt","r");


putc(c,fp4c);

}

fclose(fp4c);

fclose(fp4d);

if(getch()){

goto main;

}

case 5:


scanf("%d",&na);

printf("Change it by\t");

scanf("%d",&new_stock);



while(!feof(fp5a)){


if(na==code){

fprintf(fp5b," %d %s %f %d",code,item,price,new_stock);

}

else


}

fclose(fp5a);

fclose(fp5b);



191



while(!feof(fp5c)){



}

fclose(fp5c);

fclose(fp5d);

if(getch()){

goto main;

}

case 6:


scanf("%d",&naa);

printf("Add it by\t");

scanf("%d",&neew);



while(!feof(fp6a)){


if(naa==code){

neew=stock+neew;

fprintf(fp6b," %d %s %f %d",code,item,price,neew);

}

else


}

fclose(fp6a);

fclose(fp6b);



while(!feof(fp6c)){



}

fclose(fp6c);

fclose(fp6d);

if(getch()){

goto main;

}

case 7:

printf("Enter Code to be searched");

scanf("%d",&code_search);


printf("DATA STORED ABOUT ITEM YOUR SEARCHED\n\n");



192

printf("\n\n--------------------------------------------------------------\n");

printf("CODE\t");




while(!feof(fp7a))

{


if(code==code_search){


count++;

}

}

if(count==0){

printf("Sorry, your inquiry is not available !");

}

fclose(fp7a);

if (getch()){

goto main;

}

case 8:

printf("Enter Code_name to be searched");

scanf("%s",&item_search);


printf("DATA STORED ABOUT ITEM YOUR SEARCHED BEGINNING WITH NAME

\%s\\n\n",item_search);

printf("\n\n--------------------------------------------------------------\n");

printf("CODE\t");




while(!feof(fp8a))

{


if(strncmp(item,item_search,strlen(item_search))==0){


count++;

}

}

if(count==0){

printf("Sorry, your inquiry is not available !");

}

fclose(fp8a);

if (getch()){

goto main;



193

}

case 9:

printf("\n\n----------------------------------------\n");

printf("------------------THANKS FOR YOUR SERVICE-----%s",name);

printf("\n\n----------------------------------------\n");

if(getch()){

exit(1);

}

}

return 1;

}



194

/*THE END*/

*******


This book is an open book with an analysis of C-

Programming. In market we can only get text books

based on syllabus and curriculum with definitions

and repeated ideas. This book is an example to show

that it is important to analyze topics rather than

following the old tracks. Studying is not only to be

based upon the books but also to modify it into

different forms with different ideas according to

time and situation, which is the main theme of

understanding the topic.

Date post:	20-Jan-2016
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Programming Analysis In C (2E)

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