C programming: Session 1

7/23/2019 C programming: Session 1

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APTITUDE TRAINING

(Technical Part)(Technical Part)

At 6At 6thth Semester of Graduate ProgrammeSemester of Graduate Programme

in all Branches of Engineering / Technologyin all Branches of Engineering / Technology

Instructor : Dr. Somsubhra Gupta, IT Dept. JISCEInstructor : Dr. Somsubhra Gupta, IT Dept. JISCE

Venue: Dr. B.C. Roy Auditorium.Venue: Dr. B.C. Roy Auditorium.

Technical Aptitude Training JISCE 2015 1


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Technical SessionTechnical SessionLearning Objective:Learning Objective:

To be able to identify and address basic real world problemsTo be able to identify and address basic real world problems

To be able to solve them using elementary C programsTo be able to solve them using elementary C programs

To be conversant on Elementary Programming conceptTo be conversant on Elementary Programming concept

Session Plan:Session Plan:

C 1 C Basics C 2 If statements C 3 Loops C 4 Functions C 5 Switch-

Output

C++ 1 Structures of C++ programming C++ 2 Basic Syntax Rules

C++ 3 Elementar I/O C++ 4 Conditional Statements

Lesson Duration:Lesson Duration: 1 and our each.1 and our each.

Expected Outcome:Expected Outcome: To be able solve problems by writing codesTo be able solve problems by writing codes



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Technical SessionTechnical SessionMode of Delivery:Mode of Delivery:

u o sua roug ower o n presen a on op c w seu o sua roug ower o n presen a on op c w se

Demonstration of programmers using any C languageDemonstration of programmers using any C language compilercompiler Explanation through MarkerExplanation through Marker--White BoardWhite Board

Training Plan:Training Plan:

Discussion of Session topic in brief as everyone cleared C as a credit courseDiscussion of Session topic in brief as everyone cleared C as a credit coursein 2in 2ndnd Semester of their res ective B. Tech.Semester of their res ective B. Tech. ro rammero ramme

Writing programs on Board / Editor from previous day assignmentsWriting programs on Board / Editor from previous day assignments

Solving weekSolving week--wise (all) assignments religiously.wise (all) assignments religiously.

Encouraging content beyond assignments.Encouraging content beyond assignments.

Training Evaluation (Training Evaluation (in selective sessions):in selective sessions):

MCQ questionMCQ question

CI tec n ca test patterns quest onsCI tec n ca test patterns quest ons



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C Language Preliminaries A C program contains one or more functions

{ }: braces (left & right) to construct a block containing thestatements of a function

Every statement must end with a ;

\ is called an escape character

Other escape sequences are: \t \r \a \\ \

Exercise: Use any editor to type and then save your first program

as main.c % gcc main.c


. .


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Begin with a letter or underscore:A-Z, a-z, _

ent ers

, ,

Guarantee that east least the first 8 characters are significant (thosecome after the 8th character will be ignored) while most of C

compiler allows 32 significant characters.Example:

_abc ABC Time time _a1 abcdefghabcdefghi (may be the same as abcdefgh)

Case sensitive

auto double if static break elseint struct case entry long switch


return union do go sizeof continue


6/23

atatype

Four Data Types (assume 2s complement, byte machine)Data Type Abbreviation Size

(byte)

Range

char char 1 -128 ~ 127

~

int

int 2 or 4 -215 ~ 215-1 or -231 ~ 231-1

unsigned int unsigned 2 or 4 0 ~ 65535 or 0 ~ 232

-1 -

unsigned short int unsigned short 2 0 ~ 65535

long int long 4 -231 ~ 231-1

~ 32-

float 4

double 8



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ar a e ec arat ons

type v1,v2,v3, , vn

Example:

int i;

int j;float k;

char c;

short int x;

long int y;unsigned int z;

int a1, a2, a3, a4, a5;



8/23

, , ,Literal

Numeric literal

fixed-pointoctal O32 (= 24D) (covered later)

hexadecimal OxFE or Oxfe (=254D) (covered later)

decimal int 32

long (explicit) 32L or 32l

an ordinary integer literal that is too long to fit in an int is

also too long for long

oa ng-po nNo single precision is used; always use double for literal

Example:


1.23 123.456e-7 0.12E


9/23

, , ,Character literal (covered later)

American Standard Code for Information Interchange (ASCII)

Printable:single space 32

0 - 9 48 - 57

A - Z 65 - 90

- -

Nonprintable and special meaning chars

\n new line 10 \t tab 9\\ back slash 9 \ sin le uote 39

\0 null 0 \b back space 8

\f formfeed 12 \r carriage return 13

\ double quote 34

\ddd arbitrary bit pattern using 1-3 octal digits

\Xdd for Hexadecimal mode

\017 or \17 Shift-Ins, ^O


\04 or \4 or \004EOT (^D)

\033 or \X1B


10/23

, , ,String Literal

String is a array of chars but ended by \0

of Data Segment, so it can not be rewritten

Exam le: BCD...A B C D \0

4 chars but takes 5 byte spaces in memory

Question: I am a string takes ? Bytes


Ans: 13+1 = 14 bytes


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, , , Character literals & ASCII codes:

char x;

x=a; /* x = 97*/

Notes:

a and a are different; why?

a is the literal 97

a is an array of character literals, { a, \0} or {97, 0}

a + b +c is invalid but a+b+c = ? (hint: a = 97 in ASCII)

a + b + c = 97 + 98 + 99 = 294 = 256 + 38

in the memory

if the code used is not ASCII code, one should check out each


value of character


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n t a zat on

If a variable is not initialized, the value of variable

may be either 0 or garbage depending on the storageclass of the variable.

int i=5;

float x=1.23;char c=A;

int i=1, j,k=5;

c ar c = , c = ;float x=1.23, y=0.1;



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Each variable has a name, address, type, and value

2)scanf(%d, &x);

4)x = 200;

er e execu on o x

After the execution of (2)x

10

After the execution of (4)x 200



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Write a program to take two numbers as input data and print theirsum, their difference, their roduct and their uotient.

Problem Inputsfloat x, y; /* two items */

roblem Out ut

float sum; /* sum of x and y */

float difference; /* difference of x and y*/

float product; /* product of x and y */

float quotient; /* quotient of x dividedby y */



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Sample Problems (Contd..)

Pseudo Code:Declare variables of x and y;

Prompt user to input the value of x and y;

Print the sum of x and y;Print the difference of x and y;

Print the product of x and y;

y not equa to zero, pr nt t e quot ent o x v e y y



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#i ncl ude i nt mai n( voi d){

f l oat x, y;

pr i nt f ( Ent er t he val ue of x: ) ;scanf ( %f , &x) ;

pr i nt f ( \ nEnt er t he val ue of y: ) ;

scanf ( %f , &y) ;

sum = x + y;

pr i nt f ( \ nt he sum of x and y i s: %f , sum) ;r i nt f nt he sum of x and i s: f x+

pr i nt f ( \ nt he di f f er ence of x and y i s: %f , x- y) ;

pr i nt f ( \ nt he pr oduct of x and y i s: %f , x*y) ;

i f ( y ! = 0)

pr n e uo en o x v e y y s: , x y ;

el se

pr i nt f ( \ nquot i ent of x di vi ded by y does not exi st ! \ n) ;

return(0);


}


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ype convers onu e

char , shor t i nt

float

doubleRule #2 (doublelongunsignedint)

If either operand is double, the other is converted to

double, and the result is doubleOtherwise, if either operand is long, the other is

converted to long, and the result is long

Otherwise, if either operand is unsigned, the other isconverted to unsigned, and the result is unsigned


Otherwise, the operand must be int


18/23

Example: c: char , u: unsi gned, i : i nt , d: doubl e, f : f l oat ,

s: shor t , l : l on ,

Expression Final Data Type Explanationc s / i i nt shor ti nt , i nt / i nt , chari nt , i nt - i nt

* ,

unsi gned*unsi gned=unsi gned,

intunsigned, unsigned-unsigned=unsigned

* * . , ,

i ntdoubl e, doubl e- doubl e=doubl e

c + i i nt chari nt

c . u ar n r u e , n ou e ru e

3 * s * l l ong shor ti nt , i nt * i nt , i ntl ong, l ong*l ong



19/23

Note:

1. Conversion of int to longpreserves sign, so does

Type conversion (contd..)

short

2.Var = expr

f = d; /* round off */

i = f;

, ,

result is undetermined */

i = l; s = i; and c = i; /* may eliminate

high order bits */



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3. If a specific type is required, the following syntax may be used, called cast

operator.

Example:

= .

x = (int)f + 1;

/* the result is 3, Q: will f value be changed? */

4. Unsigned int to int:

there is not actual conversion between int and unsigned int.

Exam le: Assumin 2s com lement machine and int is 2 b tes lon

unsigned i = 65535; int j;

j = i; /* j will be 1 */


-

unsigned i = 1 + j; /* i= 65535 */


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Practice

Code

1. F n average mar s o Stu ent

2. Greatest of 3 numbers using conditional operator

3. Bi est of two numbers in C

4. Convert CM to feet and inches

5. Even or odd program in C

.

7. Swapping of two Numbers



22/23

Assignment

:

Day

1

1. Write a Program to compute the summation up to a number

. r te a program to compute actor a o a g ven num er

3. Write a program to compute sum of all the digits of a number

4. Write a program to compute whether a number is even or odd

5. Write a program to compute the mean, variance and standard



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C programming: Session 1

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