* Alejandro Colomar:

> [[
> CC += libc-co...@sourceware.org
> CC += gcc@gcc.gnu.org
> CC += libstd...@gcc.gnu.org
> ]]
>
> Hi Florian,
>
> On 2020-09-21 10:38, Florian Weimer wrote:
>> * Alejandro Colomar via Libc-alpha:
>>
>>> I'd like to propose exposing the macro 'array_length()' as defined in
>>> 'include/array_length.h' to the user.
>>
>> It would need a good C++ port, probably one for C++98 and another one
>> for C++14 or later.
>
> For C++, I use the following definition:
>
>
>       #include <cassert>
>       #include <sys/cdefs.h>
>       #include <type_traits>
>
>
>       #define is_array__(a)   (std::is_array <__typeof__(a)>::value)

Should be decltype.

> However, there are a few problems:
>
> 1) This doesn't work for VLAs (GNU extension).
>    I couldn't find a way to do it.  Maybe I should file a bug in GCC.

I do not think VLA support is critical.  C++ programmers will be used to
limited support in utility functions.

> 2) Also, this requires C++11; I don't know how to do it for older C++.
>    Again, support from the compiler would be great.

I think limited C++98 support is possible using a function template,
where the array length N is a template parameter.  To enable use in
constant expressions, you can return a type of char[N], and the macro
wrapper should then apply sizeof to the function result.

> 3) The macro can't be used in the same places as the C version,
>    because of the `({})`.
>    The `0 * sizeof(struct{...})` trick doesn't work in C++ due to:
>       error: types may not be defined in 'sizeof' expressions

For C++11, you can use a constexpr function instead of a macro.

array_length should not be a macro in current C++ modes, so that we
retain compatibility if a future C++ standard adds array_length (or
nitems) on its own.  This is not a concern for legacy C++98 mode.

>> Maybe also ask on the libc-coord list.
>
> Ok.  Added CCs.

libc-coord is not hosted on sourceware:

  <https://www.openwall.com/lists/libc-coord/2020/01/30/1>

The discussion here veered off into C++ territory anyway.

Thanks,
Florian
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