c++ - 当父类没有为自己定义宇宙飞船运算符时覆盖宇宙飞船运算符

Question

以下代码被编译器区别对待:

#include <compare>

struct A;

struct I {
    virtual std::strong_ordering operator <=>(const A&) const { 
        return std::strong_ordering::equal; 
    }
};

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

GCC 和 MSVC 都接受它，但返回错误的 Clang 不接受:

warning: explicitly defaulted three-way comparison operator is implicitly deleted [-Wdefaulted-function-deleted]
    virtual std::strong_ordering operator <=>(const A&) const = default;
defaulted 'operator<=>' is implicitly deleted because there is no viable three-way comparison function for base class 'I'
error: deleted function 'operator<=>' cannot override a non-deleted function
    virtual std::strong_ordering operator <=>(const A&) const = default;

演示:https://gcc.godbolt.org/z/WGrGTe89z

似乎 Clang 是这里唯一的一个，因为 I::operator <=>(const I&) const未定义，所以 A::operator <=>(const A&) const必须隐式删除，并且已删除的方法不能覆盖 I 中未删除的方法.其他编译器是否也有权接受代码？

Answer 1

一旦您编写类似 A a; a < a; 的代码，其他编译器也会拒绝该代码，这让我觉得 Clang 过早地拒绝了它。在 [class.compare.default] ，标准说:

A comparison operator function for class C that is defaulted on its first declaration and is not defined as deleted is implicitly defined when it is odr-used or needed for constant evaluation.Name lookups in the defaulted definition of a comparison operator function are performed from a context equivalent to its function-body.A definition of a comparison operator as defaulted that appears in a class shall be the first declaration of that function.

因为没有表达式解析为 A::operator<=>(const A&)在您的示例中，不需要定义，也不应该拒绝它，即使该函数最终总是会被删除。