functions a group of declarations and statements that is assigned a name effectively, a named...

Functions

a group of declarations and statements that is assigned a name effectively, a named statement block usually has a value

a sub-program when we write our program we always

define a function named main inside main we can call other functions

which can themselves use other functions, and so on…

Example - Square #include <stdio.h>

double square(double a) { return a*a;}

int main(void) { double num;

printf("enter a number\n"); scanf("%lf",&num);

printf("square of %g is %g\n",num,square(num));

return 0;}

This is a function defined outside main

Here is where we call the function square

Why use functions?

they can break your problem down into smaller sub-tasks easier to solve complex problems

they make a program much easier to read and maintain abstraction – we don’t have to know how a

function is implemented to use it generalize a repeated set of instructions

we don’t have to keep writing the same thing over and over

Characteristics of Functions

return-type name(arg_type1 arg_name1, arg_type2 arg_name2, …)

{function body;return value;

}

double square(double a){

return a*a;}

int main(void){

…}

Return Statement

Return causes the execution of the function to terminate and usually returns a value to the calling function

The type of the value returned must be the same as the return-type defined for the function (or a ‘lower’ type)

If no value is to be returned, the return-type of the function should be set to ‘void’

Factorials galore#include <stdio.h>

int factorial(int n){

int i, fact = 1;

for (i=2; i<=n; i++) fact *= i;

return fact;}

int main(void){ int num;

printf("enter a number\n"); scanf("%d",&num);

printf("%d!=%d\n",num,factorial(num));

return 0;}

A Detailed Example

Write a program that receives a nominator and a denominator from the user, and displays the reduced form of the number.

For example, if the input is 6 and 9, the program should display 2/3.

Example – solution (step I)#include <stdio.h>

int main(void){

int n, d;

printf("Please enter nominator and denominator: ");scanf("%d%d", &n, &d);

Calculate n’s and d’s Greatest Common Divisor

printf("The reduced form of %d/%d is %d/%d", n, d, n/gcd, d/gcd);

return 0;}

Example – solution (step II)#include <stdio.h>

int main(void){

int n, d, g;

printf("Please enter nominator and denominator: ");scanf("%d%d", &n, &d);

g = gcd(n, d);

printf("The reduced form of %d/%d is %d/%d", n, d, n/g, d/g);

return 0;}

Example – solution (step III)/* Returns the greatest common divisor of its two

parameters. Assumes both are positive. The function uses the fact that the if r = mod(y, x) then the gcd of y and x equals the gcd of x and r. */

int gcd(int x, int y){

int tmp;

while(x > 0){

tmp = x;x = y % x;y = tmp;

}return y;

}

GCD – step by stepint main(void){

int n, d, g;

printf("Please enter … : ");scanf("%d%d", &n, &d);

g = gcd(n, d);

printf("The reduced form…", n, d, n/g, d/g);

return 0;}

n d g

6 9 ---

GCD – step by stepint gcd(int x, int y){

int tmp;

while(x > 0){


}return y;

}

x y tmp6 9 ---


int tmp;

while(x > 0){


}return y;

}

x y tmp6 9 6


int tmp;

while(x > 0){


}return y;

}

x y tmp3 9 6


int tmp;

while(x > 0){


}return y;

}

x y tmp3 6 6


int tmp;

while(x > 0){


}return y;

}

x y tmp3 6 3


int tmp;

while(x > 0){


}return y;

}

x y tmp0 6 3


int tmp;

while(x > 0){


}return y;

}

x y tmp0 3 3

GCD – step by stepint main(void){

int n, d, g;

printf("Please enter … : ");scanf("%d%d", &n, &d);

g = gcd(n, d);

printf("The reduced form…", n, d, n/g, d/g);

return 0;}

n d g

6 9 3

Example – Complete Solution

gcd.c

Exercise

Input – An integer n

Output – The n’th fibonacci number

Note – Use an appropriately defined function

Solution

fibonacci_func.c

Exercise

Write a program that gets a positive integer from the user and prints all the prime numbers from 2 up to that integer.

(Use a function that returns 1 if its parameter is prime, 0 otherwise)

Solution

is_prime_func.c

Exercise Newton was the first to notice that for

any positive n, and when x0=1, the following series converges to sqrt(n) –

Use this fact to write a program that accepts a positive number and outputs its square root Hint – the thousandth element of

Newton’s series is a good-enough approximation

Solution

sqrt.c

The Great Void Sometimes there’s no reason for a function to

return a value In these cases, the function return type should

be ‘void’ If the ‘return’ keyword is used within such a

function it exits the function immediately. No value needs be specified

Calling ‘return’ in a function returning void is not obligatory

If the function receives no parameters, the parameter list should be replaced by ‘void’

Examplevoid ShowHelp(void){

printf("This function explains what this program does…\n");printf("Yadayadayada");

/* …. */}

int main(void){

char choice;

printf("Please enter your selection: ");scanf("%c", &choice);

if (choice==‘h’) ShowHelp();else if /* Program continues … */

}

Pass-by-value

Function arguments are passed to the function by copying their values rather than giving the function direct access to the actual variables

A change to the value of an argument in a function body will not change the value of variables in the calling function

Example – add_one.c

add_one – step by stepint add_one(int b){

b=b+1;return b;

}

int main(void){

int a=34,b=1;

a=add_one(b);

printf("a = %d, b = %d\n", a, b);return 0;

}

a b

34 1

Main() memory state


b=b+1;return b;

}

int main(void){

int a=34,b=1;

a=add_one(b);


}

a b

34 1

Main() memory state

b

1

add_one memory state

a b

34 1

Main() memory state


b=b+1;return b;

}

int main(void){

int a=34,b=1;

a=add_one(b);


}

b

2

add_one memory state


b=b+1;return b;

}

int main(void){

int a=34,b=1;

a=add_one(b);


}

a b

2 1

Main() memory state

Riddle me this#include <stdio.h>

int factorial(int n){

int fact = 1;

while (n>1) { fact *= n; n--; }

return fact;}

int main(void){ int n;

printf("enter a number\n"); scanf("%d",&n);

printf("%d!=%d\n", n, factorial(n));

/* What will this print? */ printf("n = %d\n", n); return 0;}

Scope of variables

A variable declared within a function is unrelated to variables declared elsewhere, even if they have the same name

A function cannot access variables that are declared in other functions

Example – scope.c

Wrong way to do itint add_one(int b){

a=b+1;}

int main(void){

int a=34,b=1;

add_one(b);


}

Function Declaration

Most software projects in C are composed of more than one file

We want to be able to define the function in one file, and to use it in all files

For this reason, the function must be declared in every file in which it’s called, before it’s called for the first time

the declaration contains: the function name the data types of the arguments (their names are

optional) the data type of the return value

Function Declaration#include <stdio.h>

int factorial(int a); /* Function Declaration! */

int main(void){ int num;

printf("enter a number\n"); scanf("%d",&num);

printf("%d!=%d\n",num,factorial(num));

return 0;}

int factorial(int a){ int i,b=1; for(i=1; i<=a; i++) b=b*i; return b;}

Function Declaration

stdio.h actually contains a large set of function declarations

The #include directive tells the compiler to insert these declarations into the file, so that these functions could be called

The math library

A collection of mathematical functions Need to include the header file math.h

(#include <math.h>) Use functions of the library, e.g.

double s,p;

s=sqrt(p); Declared in math.h :

double sqrt ( double x );

The math library

sin(x), cos(x), tan(x) x is given in radians

asin(x), acos(x), atan(x) log(x) sqrt(x) pow(x,y) – raise x to the yth power. ceil(x), floor(x) …and more

Exercise

Write a function that uses the formula

2/6=1/1+1/4+1/9+1/16+…+1/n2 (where n goes to infinity)

in order to approximate . The function should accept an argument n which determines the number of terms in the formula. It should return the approximation of .

Write a program that gets an integer n from the user, and approximate using n terms of the above formula.

Solution

pi.c

Exercise

Modify the previous function that approximates . The function should accept an argument specifying the desired accuracy, and keep adding terms until the contribution of the next term drops below this level.

Write a program that gets a (small) double, epsilon, from the user, and approximates within this function.

Solution

pi_eps.c

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