Why swapping does not occurs

Question

#include<stdio.h>
int swap(int a,int b)
{ int temp;
temp=a;
a=b;
b=temp;
return(a,b);
}
int main()
{
int c=5,d=4;
printf("after swap");
swap(c,d);
printf("%d%d",c,d);
return 0;
}

Answer 1

You can't return two numbers.

return(a,b);

Your return type is int then you must return only an int.

Answer 2

1. Change the `swap` function signature to accept pointer type for parameters instead of `int`. The function doesn't need to return anything, because it directly modifies the given arguments. So change return type to `void`.

void swap(int* a, int* b)

2. Remove the `return` statement from `swap` function. A function with `void` return type should not return anything explicitly.

3. Inside `swap` function use dereference operator '*' to obtain and/or modify values of given arguments, <a> and <b>.

int temp = *a;
*a = *b;
*b = temp;

4. In main function, pass the address of the arguments instead of their value.

swap(&c, &d);

P.S. Please consider revising C lesson chapters related to pointers 👍

Answer 3

Ragul P, there is a concept of call by value and call by reference
Please go here for better understanding.

http://cs-fundamentals.com/tech-interview/c/difference-between-call-by-value-and-call-by-reference-in-c.php