Question

I have the following code:

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

typedef void (*func_t)(void * data);

void func2(int * arg, func_t free_func) {
        free_func(arg);
}

void func(int * a) {
        printf("%d\n", *a);
}

int main(int argc, char ** argv) {
        int a = 4;
        func2(&a, func);
        return 0;
}

Compiling which gives me
warning: passing arg 2 of `func2′ from incompatible pointer type

Why is that? Shouldn’t int pointer be compatible with void pointer?

Answer 1

You can also do this:

free_func((int *)arg);

Note though that converting a void * to T* (in this case int*)
is unsafe due to the side-effects of using a T* that is not actually pointing to a T

e.g.

char i = 0;
char j = 0;    
char* p = &i;
void* q = p;
int* pp = q;    /* unsafe, legal C, not C++ */
printf("%d %d\n",i,j);
*pp = -1;   /* overwrite memory starting at &i */
 printf("%d %d\n",i,j);