When I'm writing this code:
#include <stdio.h>
int main()
{
printf("%p\n",main);
printf("%d\n",main);
return 0;
}
my compiler shows me this output:
00401318
4199192
I'm interested to know what actually is printed. I googled my question, but have found nothing. :(
Thanks in advance.
This is not well-defined.
You're using %p, which expects an argument of type void *, but you're actually passing it a value of type int (*)(), i.e. your (also badly defined) main() function.
You cannot portably cast a function pointer to void *, so your code can never be correct.
On most typicaly systems, sizeof (void *) == sizeof main, so you simply get the value interpreted as a void * which probably will simply be the address of the function.
Passing a function address to printf() with a format specifier of %d is even worse, since it's quite likely that sizeof (int) != sizeof main and then you get undefined behavior.
This is not good code.
main is a function pointer of type int(*)(void)
printf("%p\n", main);You are printing the address of that pointer, which, on your platform has been successfully cast to a void*. This will be fine if sizeof(main) == sizeof(void*).
printf("%d\n", main);This will give you undefined behaviour since %d is not a good format specifier for a pointer type.
printf is guaranteed to work correctly only if sizeof(void*) is equal to sizeof(int) on OP's platform, but it is generally considered as undefined behavior. One last thing - I'm not the one who downvoted you for this answer, so please don't go around downvoting random answers of my own.
%p formats it as hex, and %d formats as decimal. 0x401318 == 4199192%p would be equivalent to %lld (or more precisely, to %llx).
%p should be fine though". Function pointers are not compatible with void* pointers.
sizeof(void*)is equal tosizeof(int)on your platform, but it is generally considered as undefined behavior.