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I'm learning c99 and after reading about structures I found the following macro in the Linux Kernel code:

#define FIELD_SIZEOF(t, f) (sizeof(((t*)0)->f))

I.. what? Usage:

#include <stdio.h>
#define FIELD_SIZEOF(t, f) (sizeof(((t*)0)->f))

struct book {
    char title[100];
    char author[100];
};

int main(void)
{
    printf("%lu\n", FIELD_SIZEOF(struct book, title)); // prints 100
}

Here's the expansion (gcc -E)

printf("%lu\n", (sizeof(((struct book*)0)->title)));

What really baffles me is the 0. I replaced it with 1, 2, +1, -1, +999 and 'a' and "hello", and it always works.

There's no comment in the source. I know that -> is used to access a struct member through a pointer, but how can ((struct book*)0) be a pointer? How does the macro work?

Haile
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2 Answers2

6

The key here is that sizeof is calculated by the compiler at compile time. So, the pointer you specify is never actually dereferenced. After all, where an object is located won't change its size. The sizeof operator will evaluate its operand purely in terms of types.

So, the address you use is actually irrelevant, but 0 (NULL) is a common choice.

FatalError
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  • More precisely, the operand of `sizeof` *is not evaluated* (unless it's of a variable-length array type, which doesn't apply here). – Keith Thompson Oct 17 '13 at 23:19
  • Why are there parentheses in `(t*)0`? Isn't it equivalent to just `t*0` if I get it right? –  Dec 14 '16 at 02:29
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    @light2yellow: here `t` is the struct type, and hence `(t*)` is typecast to type `t *`. – FatalError Dec 14 '16 at 04:18
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    @FatalError oh, of course. Now it seems to make much more sense. But how is it possible to do the `->` stuff and why do we need a null pointer cast? How is it possible to just pass the type and get its member's size? I can imagine this operation using a pointer to our struct (allocated dynamically) but not something like this. Will this method work with dynamically allocated structs and struct members too (such as `struct foo {char **bar;};`)? Is this because the information of the needed space is always already inside every struct at compile time? –  Dec 14 '16 at 13:03
2
sizeof(((t*)0)->f)

works because sizeof is defined to not evaluate its argument (there are some exceptions when VLA are involved).

ouah
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