CODEWITHSUNDEEP
cfunction
Mukit09
Mukit09

Reputation: 3399

What is printed when I write function name in C using printf() function

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.

Upvotes: 3

Views: 337

Answers (2)

Bathsheba
Bathsheba

Reputation: 234715

main is a function pointer of type int(*)(void)

  1. 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*).

  1. printf("%d\n", main);

This will give you undefined behaviour since %d is not a good format specifier for a pointer type.

Upvotes: 4

unwind
unwind

Reputation: 399833

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.

Upvotes: 4

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