I feel like that particular point is more about the fact that new T[0]
treats the size as a runtime argument, hence the requirement that T be
default constructible, even if the default constructor isn't run. I feel
like we don't need to apply that logic to std::array, since we know the
size at compile time.

On Fri, Dec 8, 2023 at 1:04 PM Jens Maurer <jens.maurer_at_[hidden]> wrote:

>
>
> On 08/12/2023 00.27, Arthur O'Dwyer wrote:
> > On Thu, Dec 7, 2023 at 2:00 AM Jens Maurer <jens.maurer_at_[hidden] <mailto:
> jens.maurer_at_[hidden]>> wrote:
>
> > A consistency point of view:
> > T * p = new T[n];
> > (where n might be a constant or not)
> > is now required to have T default constructible even if n = 0. CWG
> 2102 <https://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#2102>.
> >
> >
> > That's a different (and perhaps even deeper) can of worms, right?
>
> Sure, but it's also about empty arrays.
>
> Jens
>
>
> > We currently lack specification for whether any of these expressions are
> supposed to work:
> > std::array<const int, 0> ca = {}; // OK
> > std::array<const int, 0> cb; // unclear; libc++ and *Microsoft*
> reject; *libstdc++* accepts
> > ca = ca; // unclear; libc++ and Microsoft reject; libstdc++ accepts
> > ca.fill(42); // unclear; libc++ and *libstdc++* reject; *Microsoft*
> accepts
> > ca.swap(ca); // unclear; libc++ and libstdc++ reject; Microsoft
> accepts
> > Is there yet an LWG issue to clarify the well-formedness of the
> "mutating" member functions of array<const T, N>?
> >
> > Thanks,
> > Arthur
>