Wrong constraint for std::optional<T>::operator=(U&&)

[jwakely]

Bugzilla Link	38638
Resolution	FIXED
Resolved on	Jun 27, 2019 11:41
Version	6.0
OS	All
CC	@dwblaikie,@mclow

Extended Description

The following should compile:

#include
constexpr int foo(int i) {
std::optional o;
o = i;
return *o;
}

Wandbox tells me:

prog.cc:2:15: error: constexpr function never produces a constant expression [-Winvalid-constexpr]
constexpr int foo(int i) {
^
prog.cc:4:5: note: non-constexpr function 'operator=<int &, void>' cannot be used in a constant expression
o = i;
^
/opt/wandbox/clang-6.0.0/include/c++/v1/optional:773:5: note: declared here
operator=(_Up&& __v)
^

https://wandbox.org/permlink/OrXOqTuQHDSddHod

The note shows that operator=(U&&) is being used, but that should not participate in overload resolution unless conjunction_v<is_scalar, is_same<T, decay_t>> is false.

In my code is_scalar is true, and is_same<int, decay_t<int&>> is true.

[jwakely]

In my code is_scalar is true, and is_same<int, decay_t<int&>> is true.

Actually, in all code is_scalar is true, I'm not special ;-)

[mclow]

Fixed in revision 364574.