为什么 std::declval 不是 constexpr？

Question

As in question - why is code like this illegal in cpp?

static_assert(std::declval<std::array<int, 4>>().size() == 4);

Is it an overlook in standard or there is some deeper rationale why std::declval is not constexpr ?

Answer 1

This line:

static_assert(std::declval<std::array<int, 4>>().size() == 4);

fails to compile, because you are using declval in an evaluated context. This is not allowed, and if you do that your program is ill-formed. declval can only be called in unevaluated contexts such as in a decltype or sizeof .

Making a function constexpr means that it can be called at either run-time, or compile-time. Since declval simply can't be called, there's no point making it constexpr . I suppose there wouldn't be any harm in making it constexpr , but either way, it doesn't matter.