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?
Why is std::declval not constexpr?
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1 Answers
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.
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