Re: Expose 'array_length()' macro in <sys/cdefs.h> or <sys/param.h>

[Jonathan Wakely <jwakely@redhat.com>]

On 21/09/20 12:33 +0200, Florian Weimer via Libstdc++ wrote:
>* Alejandro Colomar:
>
>> [[
>> CC += libc-coord@sourceware.org
>> CC += gcc@gcc.gnu.org
>> CC += libstdc++@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.

And it's wrong for references to arrays, it should be
is_array<typename remove_reference<decltype(a)>::type>::value.

>> 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.

Right, it's trivial to write in any version of C++:

template<typename T, std::size_t N>
#if __cplusplus >= 201103L
constexpr
#endif
inline std::size_t
array_length(const T(&)[N])
#if __cplusplus >= 201103L
noexcept
#endif
{ return N; }

>> 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.

A macro is 100% unacceptable for C++.

This function already exists anyway, see std::size:
https://en.cppreference.com/w/cpp/iterator/size