On 21/09/20 12:33 +0200, Florian Weimer via Libstdc++ wrote:
* 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.
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
