[expr.const] Check the result object of a prvalue

[frederick-vs-ja]

Fixes #5422.

The remain issue (modulo CWG2558) is that we shouldn't imply a prvalue is an object, which is not true since C++17.

[lprv]

A prvalue may not have a result object, at least in the scalar case.

[frederick-vs-ja]

A prvalue may not have a result object, at least in the scalar case.

Although temporary materialization doesn't happen in some cases, I think it's harmless to require the result object for the purpose of checking whether the prvalue is a constant expression.

[frederick-vs-ja]

Rebased and force-pushed. @jensmaurer @tkoeppe

[jensmaurer]

I think this is an improvement.

[lprv]

Although temporary materialization doesn't happen in some cases, I think it's harmless to require the result object for the purpose of checking whether the prvalue is a constant expression.

I disagree.

Here, for example, we surely want f() to be a constant expression, but it does not have a result object:

consteval int f() { return 42; };
int x = f() + 1;

The existing wording isn't correct but this change takes it further in the wrong direction.

[jensmaurer]

Ok. Maybe just "result"?

[frederick-vs-ja]

Although temporary materialization doesn't happen in some cases, I think it's harmless to require the result object for the purpose of checking whether the prvalue is a constant expression.

I disagree.

Here, for example, we surely want f() to be a constant expression, but it does not have a result object:

consteval int f() { return 42; };
int x = f() + 1;

The existing wording isn't correct but this change takes it further in the wrong direction.

What's the harm if we require a result object for the determination? Did you mean that if we can't even say "a prvalue whose result object..." without actual temporary materialization? But IIUC in the scalar case, addtional temporary materialization shoudn't be observable.