Использование члена int в качестве параметра шаблона внутри функции-члена consteval ⇐ C++

[Гость]

Я пытаюсь использовать

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int

member of a

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constexpr

object as a template parameter. The idea is that I create a

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constexpr Symbol

and then have that converted into a

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SymbolRef

(in the real code that 1 is different for different objects).

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template
struct SymbolRef { };

struct Symbol {
int const id_;

consteval Symbol() : id_(1) { }

template
consteval operator SymbolRef() const
{
return SymbolRef{};
}
};

However, trying to compile the above code I get the error:

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error: non-type template argument is not a constant expression
12 |     return SymbolRef{};
|                      ^~~
:12:22: note: implicit use of 'this' pointer is only allowed within the evaluation of a call to a 'constexpr' member function

The phrase "is only allowed within the evaluation of a call to a 'constexpr' member function" seems to not apply: this is a

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consteval

function!?
I understand that this code is ill-formed when

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operator SymbolRef()

would be a

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constexpr

member function (in spite of the error message!) because in that case the function must also compile without the

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constexpr

, in which case

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this->id_

can't be used as a template parameter.
Why is this code illegal when using it with

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consteval

?
EDIT:
The following does compile with g++ - but only when using

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this->id_

.

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template
struct SymbolRef { };

struct Symbol {
int const id_;

consteval Symbol() : id_(1) { }

template
consteval operator SymbolRef() const
{
return SymbolRefid_>{};
}
};

I am guessing it is a compiler bug therefore? It seems that the correct behavior is that it would compile with or without

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this->

and clang++ is just wrong.


Источник: https://stackoverflow.com/questions/781 ... r-function