Introduction to C : set 1 © Walter Milner 2005 :::: Slide :1 Starting C An introduction to...

Introduction to C : set 1 © Walter Milner 2005 :::: Slide :1

Starting CStarting C

An introduction to programming in ANSI C This material is intended to get people started in writing

C code in reasonable style It is intended to be cross-platform and not compiler-

specific However there are a few slides about getting started in

Visual Studio ( VC++ 6)


Software needed - a C compilerSoftware needed - a C compiler

· Microsoft Visual C++ or MS Visual Studio or Watcom or CodeWarrior· Turbo C v.2.01, free from http://community.borland.com/museum/ (just 3 floppies, works in DOS)· Turbo C++ v. 4.5, free on a magazine CD-ROM ( Windows)The GNU C compiler, free from http://www.gnu.ai.mit.edu/software/gcc/ .This is a command line compiler which you may find tricky to set up and use to start withIf you use Linux or another version of Unix, you will find you already have a C compiler, either installed or still on the CD.


BooksBooks

Computer Science A Structured Programming Approach Using C by Forouzan and Gilberg publishers Brooks/Cole

The C Programming Language by Kernighan and Ritchie second edition published by Prentice Hall


Using MS Visual C++ IDE - 1Using MS Visual C++ IDE - 1

1. Start MS VC++

2. Click File..New

3. Set up the dialog box like this

a console application

project name

where it is saved


Using MS Visual C++ IDE - 2Using MS Visual C++ IDE - 2

1. A wizard starts – choose ‘Empty Project’

2. Click OK

3. You will see this

select FileView not ClassView


Using MS Visual C++ IDE - 3Using MS Visual C++ IDE - 31. From the

menu, click Project.. Add to Project.. New..

2. You will see this..

We are actually adding a C source code file

A filename ending in .c


Writing the programWriting the program

to run it


A first programA first program

/* A first program */ int main() { int x, y; x = 5; y = x + 7; return 0; }


OutputOutput

/* a program with output */ #include <stdio.h> int main() { int x,y; x = 5; y = x + 7; printf(" y is %i \n ", y ); return 0; }


1. Enter the first program into an editor -/* a program with output */ #include <stdio.h> int main() { int x,y; x = 5; y = x + 7; printf(" y is %i \n ", y ); return 0; }Save it and compile it. Correct any syntax errors reported. Run the program. If you do not get the answer 12, fix it! 2. Write a program to work out the area of a rectangle. Have 3 integer variables, called width, length and area. Make the width 5 and the length 10. Calculate the area and use printf to output it. 3. Use printf to output a Christmas tree, like * *** ***** ******* *


What compilers doWhat compilers do1: int main() 2: { 00401010 push ebp 00401011 mov ebp,esp 00401013 sub esp,8 3: int x,y; 4: x = 5; 00401016 mov dword ptr [x],5 5: y = x + 7; 0040101D mov eax,dword ptr [x] 00401020 add eax,7 00401023 mov dword ptr [y],eax 6: return 0; 00401026 xor eax,eax 7: } 00401028 mov esp,ebp 0040102A pop ebp 0040102B ret


#include <stdio.h> #include <limits.h> int main() { int i; short s; long x; printf("Biggest integer = %i\

n",INT_MAX); /* get 32767 */ i = 255; x = 128000L; printf("i = %i and x = %li\n",i,x); /* get 255 and 128000 */ return 0; }

Variables must be one of a small set of data types. All data is represented internally as binary patterns, but different types are encoded into binary in different ways

The standard data type for an integer (whole number) is int

Often, an int is stored in 2 bytes = 16 bits. In that case, ints are restricted to the range ±32000

but the range of all data types in C is implementation-dependent ie not defined by C - the opposite to Java

limits.h contains the ranges of these types and will contain different values in different versions

the type short is an integer which is uses the same or less memory than int

long uses at least as much as an int. Often long uses 4 bytes = 32 bits, giving a range of about ±2 500 000 000

Integral data typesIntegral data types


#include <stdio.h>

int main()

{ float f; double d;

f = 1.23456789; printf("%f\n",f);

d = 1.2345678912345; printf("%5.13f \n",d);

f = 7 / 2; /* f becomes 3.0 */

f = 7.0 / 2.0; /* f becomes 3.5 */ }

Floating point numbers contain decimal parts Floating point is less precise and slower than

integer The basic type is float which is often stored in 4

bytes The most often used is double which is extended

precision and greater range Details of ranges and precision (implementation-

dependent) is in <float.h> The formatting sequence for a double or float in

printf is %f. A form like %5.13f indicates the number of digits before and after the decimal point. %e gives scientific notation display.

7 / 2 requests the division of 1 integer by another, so integer division is used, giving the result 3

7.0 / 2.0 divides one float constant by another, so floating point division is used, and 3.5 is the result

Floating point typesFloating point types


Arithmetic operatorsArithmetic operators The arithmetic operators are + - * / and %, which is the

remainder eg 11 % 9 = 2. Each has a precedence level so that BODMAS works and

brackets may be needed eg 4 * 2 + 2 is 14, 4 * ( 3 + 2) is 20 There are floating point and integer versions of these,

depending on the types of the operands.eg 7 / 3 gives 2

For expressions with more than one type, values are converted to the ‘biggest’ typeeg in 3.5 + 7, the 7 is converted to 7.0 and floating point addition is used.

= is also an operator, and the value is the right hand side - for example x = 7; does 2 things - it makes x to be 7 (a side-effect), and it produces a value of 7.

So you can say for example y = x = 7; Set 1-2


1. Write a program which calculates and outputs the circumference ( 2 r ) and area ( r2 ) of a circle. Use a variable called radius of type double. Take as 3.1416. Test your program works. 2. Write a program which calculates the average of 3 numbers. Use double type variables called x1, x2 and x3 as the three numbers. Check it works.


Type castingType casting An expression of one data type can be assigned to a variable of

another type such as

int x;long lo;x = 3;lo = x;

This is called a type cast Casting a value to a ‘bigger’ type, such as char to int, or float to

double, produces no data loss - but the reverse potentially can The example above is an implicit cast - it is up to the compiler to

identify that a cast is needed, and provide the code to do it An explicit cast is like this

lo = (long) x;where (long) tells the compiler to convert the type to long

Implicit casts are bad practice - they imply that the programmer is unaware of the type of the data involved

Compiler settings should be such as to generate warnings on implicit casts - then the programmer can write explicit casts where appropriate


Conditional - ifConditional - if

The idea is that the machine evaluates an expression and takes different actions depending on whether the result is true or false

Note the round brackets around the condition The { and } form a block of statements to do. If

there is only one statement, they are not needed. The else part can be missed out if it is not

needed. A value of 0 is 'false', while any non-zero value is

'true' - see later the relational and logical operators can be used,

so that if ( ( x == 4 ) || ( y < x + 7 ) ) ..means if x is 4, or y is less than x+7, do something..

if ( x > y ) { x=7; y = 55; } else { y = 9+x; printf(“Hello”); }

Relational operatorsSymbol means..

> greater than

< less than

== equal to

>= Greater than or equal to

<= less than or equal to

!= not equal to

Logical operatorsSymbol Means Idea&& AND both things true|| OR either thing true! NOT


Input - scanfInput - scanf

This is run-time input of data, normally from the keyboard For example

int i; scanf("%i", &i);execution pauses at the scanf, and waits for the user to type in a value for i

Note the &, the address operator, in front of the variable scanf is prototyped in stdio.h

scanf is not usually used in real programs (not even DOS ones) because of the difficulty with validation. In other words if the user does not type in a valid value - its too late.. but scanf is OK for initial teaching programs


1. Write a program to work out the area of a rectangle, using scanf to input data at run-time. Have 3 integer variables, called width, length and area. Calculate the area and use printf to output it. Can you find data to enter which you crash the program? 2. Write a tax calculator, which inputs a person's salary then outputs the tax, calculated by the following - if they earn less than £10 000, the tax is zeroover £10 000, they pay tax at 20% 3. Write some code to input a person's age and validate it. If the age is less than 0 or more than 120, output a message 'Age invalid' - otherwise output 'Age OK'.


Increment, decrement and assignment Increment, decrement and assignment operatorsoperators

The need to increase a variable by 1 is so common there is an increment operator - for example x++; is the same as x = x + 1 ;

Similarly there is a decrement operator x--; is the same as x = x - 1 ;

For both these, there are pre and post versions. For example, post-increment - y = x++; first makes y equal to x, then increases x by 1 -

but pre-increment y = ++x; first increases x by one, then makes y equal this new value of x

Assignment operators are short-hand. For example += x = 7; x += 9; same as x = x + 9 - so x becomes 16

More examples -Operator Example Equivalent

-= x -= 4 ; x = x - 4;

*= x *= y + 3; x = x * (y + 3);

%= x %= 2; x = x % 2;


Expressions and statementsExpressions and statements An expression is something which has a value. For example assuming the

variables have been declared and given values, all of the following are expressions -

2 + 2 5 * x + 4 4 + ( y + 3 ) / ( x + 2 ) sin( 3 * x ) /* sin is a function prototyped in math. h */ 38 x = 4 /* remember = is an operator with a side-effect */ x++ /* post-increment, side-effect, value is the initial value of x */ An expression becomes a statement if it is followed by a ; So

x + 2; is a syntactically valid statement in C - but a good compiler would warn you that 'code has no effect'

maybe you meant to type x = 2;

especially since + and = are on the same key!!!


1. Look at this piece of code - int a, b, c; a = 4;b = ++a;c = a++;b *= a;c -= 3; What should be the values of a b and c after this? Work it out on paper, then write a program to check. 2. Classify each of the following as an expression, a statement, or as invalid -

1 + 1 x = 3; x + 2

1 + 1; x++ x++;

a = b = c; x++2; 2++


if - againif - again remember 0 is ‘false’, and any non-zero value is ‘true’ programmers often use this to compress code eg

x = 3 * y + 5;if ( x == 0 )……

would probably be coded asif ( ! ( x = 3 * y + 5 ) ) …. this uses the side-effect of = to execute the assignment and the value is the value of x if x is 0, !x is 1, which is true, so the if is triggered

bug number 1..y = ….something...;x = y;if (x=0)…..many compilers will warn this


conditional - switchconditional - switch suppose you want to execute various alternative blocks of code, depending on

the value of a variable - you might write

if (x==1)…..

else if (x==2)…..

else if (x==3)…..

the switch construct caters for this switch (x)

{ 1: ….. ……

2: ….…..3: ….....default: ….…}

but.. the 1, 2 ,3 etc must be constant integral expressions when one case triggers, it falls through all other cases so the usual form is..

switch (x){

1:

….. ……

break;2:

….…..

break;

default: ….…}


conditional - the ternary operatorconditional - the ternary operator if you want to get one value if an expression is

something, and an alternative if it is not.. such as

if ( x > 38 )y = 47;

elsey = 32;

it is faster to writey = ( x > 38) ? 47 : 38 ;

which is likey is.. is x >38? if so, 47, else 38

Set 1-5


char typechar type/* char */ #include <stdio.h>

int main() { char c; c = 'a'; printf("c = %c or %i\n",c,c); /* a or 97 */ c = 'b'; c = c + 2; printf("c = %c\n",c); /* gives d */ return 0;

}

char is to be thought of as a character, or as a small integer

Traditionally, characters are represented in a machine by codes which are 1 byte long. Most PCs use the ASCII code system, in which A is 65, B is 66, C is 67,.. a is 97, b is 98,.. 0 is 48, 1 is 49, a SPACE is 32 and so on. There are other character codes which are 1 byte long, such as EBCDIC

A char variable can be assigned a value like 'a' single quotes 'like' and NOT "double quotes", which delimit strings

A char can be displayed using the %c formatting sequence in printf

Since chars are small integers, you can do arithmetic with them, like c = c + 2


1. Use switch to write a simple calculator which would run as follows (sample user input is in italics ) - Input first number: 2 Input second: 3Choose operation ( + - * or /): + Result = 5 Use scanf for input, and a char variable for the operation.

Note : if you try to input the char using scanf("%c",&c); it picks up the previous carraige return – so sayscanf(" %c",&c); // space before %c skips over CR


loops - do..whileloops - do..while use a loop if some process is to be repeated there are 3 basic loop structures in C the do while structure repeatedly executes a block of

code so long as some condition is true for example to output the numbers 1 to 10.. x = 1;

do{printf(“%i \n”, x);x++;}

while (x<11); or..

x = 1;do

printf(“%i \n”, x++);while (x<11);


a do..while loop will always execute the loop body at least once, since the condition is checked at the end. Sometimes this is inappropriate eg to delete all files in a directory and the directory is empty

an alternative loop checks the condition at the start for example to output the numbers 1 to 10..

x = 1;while (x<11)

{printf(“%i \n”, x);x++;}

or, to output 10 to 1..

x = 11;while (--x) printf("%i \n",x);

loops - whileloops - while


Loop templatesLoop templates

Task:

Output a message 10 times

This kind of process just needs an ‘index variable’ which is used to count the number of times the loop should repeat


Loop Template 1Loop Template 1

In pseudo-code

initialise counter to 0

while ( counter < 10 )output messageincrease counter by 1

In C:

#include <stdio.h>

int main(){int counter;counter = 0;while (counter<10)

{printf(“Hello\n”);counter++;}

return 0;}


Loop template 2Loop template 2

Task:

•Add up the integers from 1 to 100

•There are 2 aspects to this –•Go through the numbers 1 to 100•Add them up

•We need a counter variable to represent the 1 to 100•and another which is the “running total”



Pseudo-code plan –

Initialise counter to 1Initialise total to 0

while ( counter < ?? )add counter into totaladd 1 to counter

In C#include <stdio.h>

int main(){int counter, total;counter = 1;total = 0;while (counter<101)

{total += counter;counter++;}

printf("Total = %d\n",total);return 0;}



Task

Input integers from the keyboard, until a rogue value of –99 is enteredOutput the average

Thoughts – Input using scanf – need a ‘number’ variableWe have to add up the numbers – we need a total variableWe don’t know (at ‘compile time’) how many times we will loopWe have to count them – we need a counter variableWe must avoid adding in and counting the -99


Loop template 3Loop template 3Pseudo-code

Initialise total to 0Initialise counter to 0

Input number

while ( number not –99 )Add number to totalIncrement counterInput number

average = total / counterOutput average

#include <stdio.h>

int main(){int counter, total, number;double average;counter = 0;total = 0;scanf("%d",&number);while (number!= -99)

{total += number;counter++;scanf("%d",&number);}

average = (double)total / (double) counter;printf("Average = %lf\n",average);return 0;}


1. Here is a program which outputs the integers 1 to 5 - #include <stdio.h>int main(){

int stepper;stepper = 1;while (stepper<6){

printf("%d ",stepper);stepper++;

}return 0;

} We will use this as a basis for several more programs. Type it in and run it. 2. Change the program so it outputs integers from 1 to 99 (inclusive).3. Change it so it outputs from -99 up to 99.4. Change it so it goes up in steps of 2 ( so you get -99 -97 -95........97 99 ).5. Change it so it starts at 100 and goes down to 1.


1. Use a do-while loop to output the even numbers from 2 to 100 inclusive. 2. Input an integer then output 'Hello' that number of times. Include the possibility of 0 iterations. 3. Use a loop to output the sequence 3, 6, 9, 12,..300 4. Add up the integers from 1 to 100. 5. The user inputs a sequence of integers, terminated by a 'rogue value' of -99. Add up the sequence and output the result -Sample input 1, 2, 3, -99Output 6


ArraysArrays

7 13 42 4 17 22 7 66 3 9

0 1 2 3 4 5 6 7 8 9

An array is a set of data values – not just one

Think of an array as a row of boxes

Each box is an element

The elements have an index – a label for which box is which

The 10 elements

index of each element

array of 10 integers


An array is a data structure – a way of arranging a set of data items

An array is a set of numbered boxes, each box holding a value

an array marks[20] could hold the exam marks of 20 students

This would be declared asint marks[20]; < must use square brackets

The boxes are numbered 0 to 19 Some examples are

marks[0] = 35; store 35 into the first boxmarks[9] = 48; store 48 into the 10th. boxx = marks[0] + marks[1]; add up first 2 boxesk = 34;marks[k] = 17; put 17 into 35th. Boxmarks[20] = 23; !! Array bounds error !!

ArraysArrays


To process arrays loops are often used to go through the different elements

For example, this would declare an array, read values in from the keyboard, then print them out again..int index = 0;int numbers[5];while ( index < 5 ) { printf("Enter a number.."); scanf("%i", &numbers[index] ); index++;

}

index = 0;while ( index < 5 ) { printf(“ %i %i \n ", index, numbers[ index ] ); index++; }

Arrays and loopsArrays and loops


Arrays - count zerosArrays - count zerosSuppose we have an array called numbers, containing 5 integers, and we want to count how many zeros there are -

zeroCount = 0;

index = 0;

while ( index < 5 )

{

if ( numbers [ index] == 0 )

zeroCount++;

index++;

}


Arrays - find largest elementArrays - find largest element

biggest = 0;

index = 0;

while ( index < 5 )

{

if ( numbers [ index] > biggest )

biggest = numbers[index];

index++;

}


Arrays - find totalArrays - find total

total = 0;index = 0;

while (index < 5 )

total = total + numbers[index++];


Arrays of charArrays of char Strings in C are stored as sequences of character codes,

and are terminated by a 0 - ASCIIZ so “ABC” is stored in 4 bytes, with values 65 66 67 0 this can be done using arrays eg

char word[4] = “ABC”; or

char word[4];word[0]=‘A’;word[1]=‘B’;word[2]=‘C’;word[3]=‘\0’; not word[3]=‘0’;

orchar word[] = “a longer string”;

the problem with this is an array is fixed in length, so you can’t store longer strings in the array. Consequently pointers to chars are more often used..


1. Fill an array with 5 random integers ( use the function rand() from stdlib.h ). Print them out, together with the average of the 5 numbers. 2. Fill an array with 50 random integers. Print them out, then reverse them, and print them out again


for loopsfor loops2 kinds of C loops – while( ) and do..while( )

The 3rd and last kind is a for loop

It looks like –

for (…………..){....}

There are 3 parts in this bracket

This is the loop body – the statements to repeat


for loopsfor loops

for ( x=0 ; x<5 ; x++ ){....}

Initialise

change at the end of each repeat

loop repeats so long as this is true

so this loops with x having the values

0 1 2 3 and 4


for loop – output 1 to 100 for loop – output 1 to 100 (nearly)(nearly)

..

int x;

..

for ( x = 0; x < 100; x++)

printf(“%d\n”, x);


for loop – counting downfor loop – counting down

..

int x;

..

for ( x = 100; x > 0; x = x - 2)

printf(“%d\n”, x);


for loop – totalling an arrayfor loop – totalling an array

int data[100]

int index, total = 0;

..

for ( index = 0; index < 100; index++ )

total = total + data[index];


nested loopsnested loopsint x;int y;y = 0;

..

for (x = 0; x<3; x++)printf(“%d ”, x + y );

for (y = 0; y<30; y = y + 10 ) {

printf(“\n”); }

0 1 2

0 1 2

10 11 12

20 21 22


1. Use a for loop to output 1 to 100 inclusive

2. Use a for loop to output the 4 times table like

1 X 4 = 4

2 X 4 = 8

..

12 X 4 = 48

3. Modify the program in 2 so it produces all the times tables from 2 to 12


Long pieces of code are very hard to manage It is more effective to structure programs into small pieces or

units All C programs are made of units called functions main() is a special function - where execution starts - so far

the only function in the program Real programs are broken up into several functions A function should not have more than around 30 lines of code

- or it should be broken up into other functions This modularity means that

small functions are easier to write they are easier to debug they can be tested separately faster they will not stop each other working they can be written by teams of programmers they can be re-used in other projects

Functions - ideasFunctions - ideas


Function call sequenceFunction call sequence

void func() { .. return; }....int main() { .. func(); .. }

the definition of function

func

a call of func()


function examplefunction example#include <stdio.h>

void greeting() { printf("Hello\n"); return; }

int main() { greeting(); return 0; }

void means the function does not ‘return a value’

it just does something


•This works –#include <stdio.h>

void greeting(); // function prototype

int main() { greeting(); return 0; } void greeting() { printf("Hello\n"); return; }•header files (like stdio.h) are mostly prototypes

function prototypesfunction prototypes•This doesn’t work –#include <stdio.h>int main() { greeting(); return 0; } void greeting() { printf("Hello\n"); return; }•because the compiler works through program top to bottom, and•it sees the call to greeting() before it has seen the definition of what greeting is


function parametersfunction parametersA value can be sent into a function to specify what it does

This is called a function parameter#include <stdio.h>

void greeting(int howManyTimes){ int i; for (i=0; i<howManyTimes; i++) printf("Hello\n"); return;}

int main(){ greeting(5); return 0;}


functionfunction with a parameter with a parameter#include <stdio.h>

void table( int );

int main(){

table(5); return 0;}

void table( int n){

int i;for (i=1; i<13; i++)

printf("%i X %i = %i\n", i,n,i*n);return;

}

output…1 X 5 = 52 X 5 = 103 X 5 = 154 X 5 = 205 X 5 = 256 X 5 = 307 X 5 = 358 X 5 = 409 X 5 = 4510 X 5 = 5011 X 5 = 5512 X 5 = 60


global and local variablesglobal and local variablesint g;

void func()

{

int x;

..

}

int main()

{

int y;

..

}

This are local or automatic variables

This is a global or external variable


External variablesExternal variables this program is supposed to fill an

array with numbers, display them, reverse them, and display them again

so each function needs access to the array, and fill() and reverse() need to alter its contents

This can be done by making it an external variable

These are declared outside functions – they are ‘global variables’ – contrasted with those declared at the start of a function, which are automatic variables

External variables provide for unrestricted interaction between functions

In particular a function can alter them in invalid or accidental ways so..

External variables should be avoided if at all possible

int numbers[20];

void fill(void);void reverse(void);void display(void);

int main(){

fill();display();reverse();display();return 0;

}

void fill(){ /* define fill */

return;}

void display(){ /* define display */

return;}

void reverse(){ /* define reverse */

return;}


ScopeScope The scope of a variable is the section of the

program within which the variable can be accessed

The scope of an automatic variable (local variable) is the function within which it is declared

The scope of an external variable is from where it is declared to the end of the file

Automatic variables with the same name in different functions are unrelated


#include <stdio.h>int ex = 0;void foo1()

{int a;a = 1;return ;}

void foo2(){int a;ex = 3;a = 3;return;}

int main(){ex = 7;foo1();foo2();return 0;}

ex is external and has this scope.

It can be accessed in foo1(), where it will have value 7 (set in main() )

Workshop 12

Scope exampleScope example


functionsfunctions returning values returning values

This program uses a function to calculate the average of two numbers – it has to return the answer

The protoypedouble average(double, double );tells the compiler the return type, and the number and type of the arguments

The prototype must be consistent with the header of the function definition

The names of the arguments in the definition ( x and y ) are matched with the values in the function call ( a and b ) in order

#include <stdio.h>

double average(double, double );

int main(){

double a,b,c;a=1.0;b=2.0;c=average(a,b);printf("%f\n",c);

return 0;}

double average(double x, double y){

double r;r = (x+y)/2.0;return r;

}


call by valuecall by valuevoid foo( int );

int main(){

int i=1;foo(i);/* how big is i now? */

return 0;}

void foo( int n){

n++;/* n is 2 - but what is i ? */return;

}

parameters are always ‘called by value’

this means that a copy of the value of a parameter is passed into the function

the function can alter the value of the copy, but not the original


FunctionsFunctions

1. Write a function which finds the average of 3 numbers rather than 2. Write a main() to test it.

2. Write a function bigger that takes 2 integer parameters, and returns the larger one.

3. Write a function IsUpper, which takes a char argument, and

returns 1 if the argument is a capital letter, and 0 otherwise. For example,

isUpper('A') returns 1isUpper('x') returns 0hint: if (c>='A' && c<='Z') … then c is an upper case character.


1. Suppose you want to find the square root of x. Here is a method -

set y to 1

repeat

replace y by (y + x / y) / 2

y will get closer and closer to the square root of x. Write a function called root using this method.

2 The factorial of a number, written n!, is the product of all the integers up to n. So for example 4! = 4 X 3 X 2 X 1 = 24. Write a void function called PrintFactorial which has an integer argument, and which outputs the factorial of it. Test it with 10! = 3 628 800

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