Function Pointer and GCC warning

Question

Hi,

I'm currently learning how to write shellcode and get a GCC warning "assignment of incompatible pointer type". Why?

Greetings, LM

char shellcode[] = 
  "\x31\xc0"
  "\xb0\x01"
  "\x31\xdb"
  "\xcd\x80";


int main() {
  void (*func)() = shellcode;

  func();

  return 0;
}

Answer 1

Because char[] is incompatible with void(*)()!! What on earth would that even mean?

Answer 2

There is no implicit conversion of object pointers to function pointers. You'll need an explicit cast (void(*)()) (and even that isn't guaranteed to work, or compile).

Answer 3

I'll write my suggestion as an answer and whore for some reputation :-)

You need to explicitly cast the data pointer into a function pointer. With a typedef for the appropriate function pointer, you can do so without making the code too hard to read:

char shellcode[] = 
    "\x31\xc0"
    "\xb0\x01"
    "\x31\xdb"
    "\xcd\x80";

typedef void (*SHELLCODE)(void);

int main() {
    SHELLCODE func = (SHELLCODE) shellcode;
    func();

    /* ...and this will work as well, which some might find more readable: */
    ((SHELLCODE) shellcode)();

    return 0;
}

This compiles cleanly with clang 1.1, but GCC 4.4.3 will still give you a warning about the pointer conversion -- although now it is a more precise warning:

$ gcc -Wall -Wextra -ansi -pedantic shellcode.c 
shellcode.c: In function ‘main’:
shellcode.c:10: warning: ISO C forbids conversion of object pointer to function pointer type
shellcode.c:14: warning: ISO C forbids conversion of object pointer to function pointer type

Skipping the "-pedantic" flag will make it compile cleanly with GCC too, but who ever does that seriously?