This may be a bug in our optional implementation.

Given:

#include
#include <type_traits>
#include

struct S {
constexpr S() : v_(0) {}
constexpr S(int v) : v_(v) {}
constexpr S(const S &rhs) : v_(rhs.v_) {} // not trivially copyable
constexpr S( S &&rhs) : v_(rhs.v_) {} // not trivially moveable
int v_;
};

int main(int, char**)
{
static_assert (!std::is_trivially_move_constructible_v, "");
constexpr S s1(3);
constexpr S s2 = std::move(s0);
(void) s2;

constexpr std::optional<S> o1;
constexpr std::optional<S> o2 = std::move(o1);  // not constexpr

Better test:

#include
#include <type_traits>
#include

struct S {
constexpr S() : v_(0) {}
constexpr S(int v) : v_(v) {}
constexpr S(const S &rhs) : v_(rhs.v_) {} // not trivially copyable
constexpr S( S &&rhs) : v_(rhs.v_) {} // not trivially moveable
int v_;
};

constexpr bool test()
{
std::optional o1{3};
std::optional o2 = std::move(o1); // not constexpr
return o2.has_value();
}

Eric pointed out that you can't have a constexpr move-assignment operator for the non-trivial case, because it has to do a placement new (unless the optional that you are constructing from is empty)

That's why my second test fails.
Apparently libstdc++ is happy with this:

constexpr bool test()
{
std::optional o1; // empty
std::optional o2 = std::move(o1);
return o2.has_value();
}

Test fixed in revision 359162 - now the test is move-constructing a non-empty optional.