On Wed, 19 Feb 2020 at 11:17, Marc Glisse wrote:
>
> On Wed, 19 Feb 2020, Klaus Doldinger wrote:
>
> > Am 19.02.20 um 11:02 schrieb Jonathan Wakely:
> >> On Wed, 19 Feb 2020 at 07:53, Klaus Doldinger
> >> <klaus.doldinger64@xxxxxxxxx> wrote:
> >>>
> >>> The following was possible in v9:
> >>>
> >>> template<auto N>
> >>> struct A {
> >>> constexpr auto size() const {
> >>> return N;
> >>> }
> >>> };
> >>>
> >>> template<typename A>
> >>> void foo(const A& a) {
> >>> constexpr auto s = a.size();
> >>> }
> >>>
> >>> int main() {
> >>> A<10> x1;
> >>> foo(x1);
> >>> }
> >>>
> >>> In v10 this is rejected.
> >>>
> >>> Looks like a bug?
>
> Clang also rejects it.
>
> >> No, I don't think so. The parameter a is not a constant expression, so
> >> a.size() is not a constant expression, so can't be used to initialize
> >> a constexpr variable.
>
> Why not make size() static?
>
> > That would be weird.
> > If so, the follwing would be impossible:
> >
> > std::array<int, 10> a;
> > std::array<int, a.size()> b;
>
> You can still use decltype and tuple_size to access that information.
And it works if 'a' is usable in a constant expression, e.g. because
it has static storage duration.
It doesn't work if it's an automatic variable, or if you pass it by
reference to a non-constexpr function, which is the problem in the
OP's original example.
