Macro for counting number of elements in an array

Question

We all know the classic version of counting the number of elements in a C array: sizeof(a)/sizeof(*a)

But this is dangerous, because if used on a pointer it will return garbage values.

So I made this macro, utilizing gcc extensions. As far as I can tell, it works the way it is supposed to, which is causing compiler error when used on a pointer instead of array.

#define COUNT(a) (__builtin_choose_expr( \
 __builtin_types_compatible_p(typeof(a), typeof(&a[0])), \
 (void)0, \
 (sizeof(a)/sizeof(a[0]))))

/* Only to demonstrate how to use the macro. It's not part of the code
 I want reviewed. */
int main(void)
{
 int arr[5];
 int *p;
 int x = COUNT(arr);
 // This line will yield the compiler error:
 // error: void value not ignored as it ought to be
 int y = COUNT(p);
}

Any suggestions? Have I missed anything crucial?

• chux: klutt, Why int x and not size_t x, the type returned by the division?
• chux: See also Array-size macro that rejects pointers.
• Mast: Why are you trying to write the value of a void on the last line of your main? What are you trying to illustrate here?
• klutt: @chux-ReinstateMonica See comment. Also, I should have realized that a linux kernel developer already thought of this :)
• klutt: @Mast Because that will fail. I want a compiler error if it is used on a pointer instead of an array.

Answer 1

We need to protect a when it's an expression - we've put parens around a[0] where we should have put them around a:

(sizeof(a)/sizeof(a)[0])

For the same reason, I think that the second typeof should be

typeof(&(a)[0])

I have to admit that I can't easily contrive an expression of array type where the parens make a difference, but better safe than sorry...