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 = temp;
   // Gives 'Non-object type 'int (int)' is not assignable
 }

 swap2(&funcA, &funcB);

EDIT

More data here as to intention -- Some answers have been provided below which do work such as creating the function ptr using typedef, pointing them to the functions and switching those, which lets you invoke the new swapped ptrs successfully.

BUT calling the functions by their original names after swapping shows no change. Essentially I'm looking for a c equivalent of the objc "swizzle".

I'm beginning to think this isn't possible, due to c's complete lack of reflection, and would require actually modifying the binary itself (obviously not feasible). D:

Comments welcome.

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 <stdio.h>
#include <stdlib.h>

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.

Answer 2

I'm pretty sure you need pointers to function pointers to swap pointers, no? This type of swapping function swaps values; you really want to deal in addresses. The example function call wouldn't really work because C doesn't treat functions as first-class variables so you can't actually swap functions directly; you need to use pointers to function addresses, since addresses CAN be swapped:

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

int(*func1)(int) = &foo;
int(*func2)(int) = &bar;

swap2(&func1, &func2);