Is it possible to swap C functions?

Question

Looking to see if anyone knows if its possible to swap C functions...?

 void swap2(int(*a)(int), int(*b)(int)) {
   int(*temp)(int) = a;
   *a = *b;
   *b =          


        

Answer 1

If you use the function pointers like below, it is yes

typedef int (*func_pt)(int);

func_pt a, b;

void swap(func_pt * a, func_pt * b)
{
    func_pt tmp = *b;
    *b = *a;
    *a = tmp;
}

swap(&a, &b);

Or you use it as this, I think it is no:

int test1(int a)
{
    return a;
}

int test2(int b)
{
    return b;
}

swap(&test1, &test2);

Complete compiling working program

#include 
#include 

typedef int (* func_pt)(int);

func_pt a, b;

int test1(int a)
{
    printf("test1\n");
    return 1;
}

int test2(int a)
{
    printf("test2\n");
    return 2;
}

void swap(func_pt * a, func_pt * b)
{
    func_pt tmp = *b;

    *b = *a;
    *a = tmp;
}

int main(void)
{
    a = &test1;
    b = &test2;

    printf("before\n");
    a(1);
    b(1);

    swap(&a, &b);

    printf("after\n");
    a(1);
    b(2);

    return 0;

}

Output:

before
test1
test2
after
test2
test1

Some people do not try it by themselves, just say it absurd.So I give you a example.