Is reduction of `constexpr` object lifetime legal in C++?

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For ordinary objects (even for const ones), it is permissible to end their lifetime by explicitly calling the destructor. Later, for example, the program can start another object lifetime in the same memory location using placement new.

But is it legal to call the destructor of a constexpr object? Can it result in some useful or at least well-formed program?

It is easy to imagine the opposite situation:

struct A{
    int v = 0;
    constexpr void foo() { v ? throw 1 : ++v; }
    constexpr ~A() { foo(); }
};

constexpr A y;

int main() { y.~A(); y.~A(); }

This (most probably ill-formed) program is accepted by all compilers without any single warning: https://gcc.godbolt.org/z/aqMbfjxKT

And in Clang it finished by throwing an exception from constexpr destructor of A.

explicit-destructor-call constexpr language-lawyer c++
1 Answers

[dcl.constexpr] A constexpr specifier used in an object declaration declares the object as const.

You can do with such an object whatever you can do with any other const object. constexpr, apart from implying constness of the declared object, doesn't seem to affect anything about the object itself. It only affects static properties of the program, that is, which expressions are constant expressions, which template instantiations are valid, etc. There are no special provisions for run-time behaviour of objects declared with the constexpr specifier.

The Standard uses phrases similar to "constexpr object" a couple of times but it seems to be a misnomer. In my opinion it should have referred to constexpr variables instead. There should be no such thing as "constexpr objects" (like there are no inline objects), only constexpr variables (like there are inline variables). After all, constexpr is a decl-specifier, like inline. The objects that these variables refer to are just const.

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