[Solved] Why isn’t this swap program working in C?

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The void is a sub-process, it must be enclosed in braces and must not end in a semicolon.
on the other hand as the community says int * temp = * p1; is a constaint violation, it must be int temp = * p1;

Nor was exchanging values, it showed only as when they were entered.

Finally, the main returns an int, you must specify that you return a 0, that is done with return 0;

The whole program would be like that

#include <stdio.h>

void swap (int *p1, int *p2){
    int temp = *p1;
    *p1 = *p2;
    *p2 = temp;
}

int main ()
{
    int x = 10;
    int y = 20;
    swap (p1, p2);

    printf ("x: %d, y: %d\n", x, y);
    return 0;
}

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solved Why isn’t this swap program working in C?