具有默认实现的虚拟`operator <=>`是否在C ++ 20中增加了一个虚拟方法？

Question

如果使用默认实现制作 virtual C++20 three-way 比较运算符 <=>，则编译器会默默地制作一个虚拟 operator==，因为可以通过在 child class:

中覆盖它来显示

struct A { 
    virtual std::strong_ordering operator <=>(const A &) const = default; 
};
struct B : A {
    virtual bool operator==(const A&) const noexcept override;
};

所有三个主要编译器都以这种方式运行，演示：https://gcc.godbolt.org/z/nvaf94M51

它是标准强制要求的还是只是一种常见的实施实践？

附带一提，在 C++20 中我们还可以创建 consteval 立即虚函数，如果扩展该示例：

#include <compare>

struct A { 
    virtual consteval std::strong_ordering operator <=>(const A &) const = default; 
};

struct B : A {
    virtual consteval bool operator==(const A&) const noexcept override { return false; }
};

int main() {
    static constexpr B b;
    static constexpr const A & a = b;
    static_assert( (a <=> a) == 0 );
    static_assert( a != a );
    return (a <=> a) == 0; 
}

然后它给现代编译器带来了一定的困难：https://gcc.godbolt.org/z/Gsz39Kr5b

Clang 崩溃，GCC 生成一个程序，该程序在执行过程中失败。这些只是一些编译器错误，还是程序格式不正确？

Answer 1

Is it mandated by the standard or just a common implementation practice?

这是强制性的。 [class.compare.default]/5:

If the member-specification does not explicitly declare any member or friend named operator==, an == operator function is declared implicitly for each three-way comparison operator function defined as defaulted in the member-specification, with the same access and function-definition and in the same class scope as the respective three-way comparison operator function, except that the return type is replaced with bool and the declarator-id is replaced with operator==.

[Note 2: Such an implicitly-declared == operator for a class X is defined as defaulted in the definition of X and has the same parameter-declaration-clause and trailing requires-clause as the respective three-way comparison operator. It is declared with friend, virtual, constexpr, or consteval if the three-way comparison operator function is so declared. ... — end note]

[Example 1:
template<typename T> struct X {
 friend constexpr std::partial_ordering operator<=>(X, X) requires (sizeof(T) != 1) = default;
 // implicitly declares: friend constexpr bool operator==(X, X) requires (sizeof(T) != 1) = default;

 [[nodiscard]] virtual std::strong_ordering operator<=>(const X&) const = default;
 // implicitly declares: [[nodiscard]] virtual bool operator==(const X&) const = default;
};
— end example]

具有默认实现的虚拟`operator <=>`是否在C ++ 20中增加了一个虚拟方法？

Does virtual `operator <=>` with default implementation make one more virtual method in C++20?

c++

comparison-operators

language-lawyer

c++20