Why does this code not swap the numbers? [duplicate]

Answer 1

This is because you pass variables by copy and not by pointer. In other words, your swap() functions receives its own private copies of x and y and swap them and the result of swap is not visible by the caller. The correct code might look something like this:

#include 

void swap(int *a, int *b);

int main(void)
{
    int x = 1;
    int y = 2;
    swap(&x, &y);
    printf("Now x is %d and y is %d\n", x, y);
    return 0;
}

//function definition of swap
void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}