Understanding macro measuring array count from glut tutorial

Question

I am learning/reading a bit of openGL. I am currently following the tutorials from http://www.arcsynthesis.org/gltut/ . The program has a macro which is used for array length. I know array length is counted as (sizeof( array ) / (sizeof( array[0] ).

But the code has some additional stuff so the final macro is :

#define ARRAY_COUNT( array ) (sizeof( array ) / (sizeof( array[0] ) * (sizeof( array ) != sizeof(void*) || sizeof( array[0] ) <= sizeof(void*))))

I am unable to understand why is it being multiplied by a bool.

Indeed. But it's non-robust, and it's arguable that returning 0 is not helpful. — Oliver Charlesworth
– Oliver Charlesworth, Commented Aug 17, 2014 at 13:40
Yeah, it seems pretty pointless. If you have a char *, sizeof(arr[0]) <= sizeof(void *) must be true, so it will still count. Considering this is C++, you can write a constexpr function to achieve the same thing, or a regular one to have the result at runtime. Of course you could also just use std::array or boost::array. — Qaz
– Qaz, Commented Aug 17, 2014 at 13:41

Jarod42 · Accepted Answer · 2014-08-17 14:27:39Z

1

The addition is to check some bad usages of pointer instead of array as:

long long buf[42];
long long *p = buf; 

ARRAY_COUNT(p); // this produces a compilation error : division by zero.

but it fails to detect some bad usages as:

char buf[42];
char *p = buf; 

ARRAY_COUNT(p); // this gives unexpected result.

The check cannot detect bad usage for types which have a size less or equal than a pointer.

The C++ way to do that is:

template <typename T, std::size_t N>
constexpr std::size_t ArraySize(T (&)[N]) { return N; }

edited Aug 17, 2014 at 14:27

answered Aug 17, 2014 at 14:15

Jarod42

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