How to format a function pointer?

The only legal way to do this is to access the bytes making up the pointer using a character type. Like this:

#include <stdio.h>

int main() {
    int (*funcptr)() = main;
    unsigned char *p = (unsigned char *)&funcptr;
    size_t i;

    for (i = 0; i < sizeof funcptr; i++)
    {
        printf("%02x ", p[i]);
    }
    putchar('\n');

    return 0;
}

Examining the bytes of the function pointer with an lvalue of type void *, or any non character type, is undefined behaviour.

What those bytes making up the function pointer actually mean is implementation-dependent. They could just represent an index into a table of functions, for example; or they could even be the first N characters of the function's name which is looked up in the symbol table when you call through the function pointer. The only operations that need be supported on a function pointer are calling the function through it and comparison against another function pointer or NULL for strict equality/inequality, so there is very wide latitude available in how they are implemented.

There's the use of unions that can get around the warning/error, but the result is still (most likely) undefined behavior:

#include <stdio.h>

int
main (void)
{
  union
  {
    int (*funcptr) (void);
    void *objptr;
  } u;
  u.funcptr = main;

  printf ("%p\n", u.objptr);

  return 0;
}

You can compare two function pointers (e.g. printf ("%i\n", (main == funcptr));) using an if statement to test whether they're equal or not (I know that completely defeats the purpose and could very well be irrelevant), but as far as actually outputting the address of the function pointer, what happens is up to the vendor of your target platform's C library and your compiler.