How to print the address of a function?

Question

I let gcc compile the following example using -Wall -pedantic:

#include <stdio.h>

int main(void)
{
  printf("main: %p\n", main); /* line 5 */
  printf("main: %p\n", (void*) main); /* line 6 */

  return 0;
}

I get:

main.c:5: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘int (*)()’
main.c:6: warning: ISO C forbids conversion of function pointer to object pointer type

Line 5 made my change the code like in line 6.

What am I missing to remove the warning when printing a function's address?

Answer 1

This is essentially the only portable way to print a function pointer.

size_t i;
int (*ptr_to_main)() = main;
for (i=0; i<sizeof ptr_to_main; i++)
    printf("%.2x", ((unsigned char *)&ptr_to_main)[i]);
putchar('\n');

Answer 2

This whole idea is indeed non-portable, since some systems use different sized pointers to code and data.

What you really need is platform-specific knowledge of how big a function pointer is, and a cast to an integral type of that size. Unfortunately, I don't think anyone has standardized a intfuncptr_t analagous to intptr_t which can hold any data pointer.

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As R. notes in his answer, you can always treat the pointer as an array of (possibly signed or unsigned) char, this way you don't need any integral type of the correct size.