Am 19.02.20 um 13:43 schrieb Jonathan Wakely:
> 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.
Well, the following works:
A<10> x1;
A<x1.size()> x2;
std::array<int, 10> a;
std::array<int, a.size()> b;
Here x1 is not a constant expression, and it is accepted. Why?
Please explain what the difference is to what in function foo() happens.
If function foo() is changed to
template<typename T>
void foo(T a) {
constexpr auto s = a.size();
}
it is also expected.
Thats not consistent I would say.
