https://gcc.gnu.org/bugzilla/show_bug.cgi?id=87016
--- Comment #4 from bobogu at atlas dot cz --- (In reply to Jonathan Wakely from comment #3) > (In reply to bobogu from comment #2) > When you assign an int to optional<int> it is equivalent to constructing a > temporary optional<int> and then assigning that to the left-hand operand. I feel stupid, but I still don't understand how that can work in a constexpr context when the current c++17 standard doesn't specify any constexpr assignment operator. The second line of this snippet std::optional<int> bar = 3; bar = std::optional<int>(10); would call the 3rd assignment operator (https://en.cppreference.com/w/cpp/utility/optional/operator%3D), right? And the operator isn't constexpr, but works in a constexpr function.
