What type is a function name in C? [duplicate]

Question

I always understood that in C, func and &func were equivalent. I assume they should both be of type pointer, which is 8 bytes on my Win64 system. However, I just tried this:

#include <stdio.h>  int func(int x, int y) {     printf("hello\n"); }  int main() {     printf("%d, %d\n", sizeof(&func), sizeof(func));     return 0; } 

And expecting to get the output 8, 8 was surprised to get 8, 1 instead.

Why is this? What type exactly is func? It seems to be of type char or some equivalent.
What is going on here?

I compiled this with gcc -std=c99 if it makes a difference.

Answer 1

What type is a function name in C?

A function name or function designator has a function type. When it is used in an expression, except when it is the operand of sizeof or & operator, it is converted from type "function returning type" to type "pointer to a function returning type". (This is specified in C99, 6.3.2.1p4).

Now

sizeof(func)

is not valid C as sizeof is not allowed with an operand of function type. This is specified in the constraints of the sizeof operator:

(C99, 6.5.3.4p1 Constraints) "The sizeof operator shall not be applied to an expression that has function type or an incomplete type, to the parenthesized name of such a type, or to an expression that designates a bit-field member."

But

sizeof(func)

is allowed in GNU C.

There is a GNU extension in GNU C that allows it and in GNU C sizeof with an operand of function type yields 1:

6.23 Arithmetic on void- and Function-Pointers

[...] sizeof is also allowed on void and on function types, and returns 1.

http://gcc.gnu.org/onlinedocs/gcc/Pointer-Arith.html