On Wed, 19 Feb 2020 at 10:10, Klaus Doldinger
<klaus.doldinger64@xxxxxxxxx> 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?
> >
> > 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.
> >
>
> That would be weird.
> If so, the follwing would be impossible:
>
> std::array<int, 10> a;
> std::array<int, a.size()> b;
That's not at all equivalent.
A function with the constexpr specifier doesn't mean it can always be
used in a constant expression, only that it can be used in a constant
expression when called with constant arguments (including the implicit
'this' argument for a member function).
In your case you are trying to use a.size() inside a non-constexpr
function, with a non-constant argument. That means a.size() is not a
constant. So can't initialize a constexpr variable.
