为什么我可以使用默认的< =>调用==但不是用户提供的？ [英] Why can I invoke == with a defaulted <=> but not a user-provided one?

问题描述

#include <compare>

struct A
{
    int n;
    auto operator <=>(const A&) const noexcept = default;
};

struct B
{
    int n;
    auto operator <=>(const B& rhs) const noexcept
    {
        return n <=> rhs.n;
    }
};

int main()
{
    A{} == A{}; // ok
    B{} == B{}; // error: invalid operands to binary expression
}

的无效操作数，用clang-10编译为 clang -std = c ++ 20 -stdlib = libc ++ main.cpp

compiled with clang-10 as clang -std=c++20 -stdlib=libc++ main.cpp

为什么 A {} == A {} 工作，但不是 B {} == B {} ？

推荐答案

在飞船运算符的原始设计中，允许 == 调用< ; => ，但是后来出于效率考虑而被禁止使用（< => 通常是实现 == ）。 operator< =>（）=默认值仍被定义为隐式定义 operator == ，它会正确调用<$为方便起见，在成员上使用c $ c> == 代替< => 。因此，您想要的是这样：

In the original design of the spaceship operator, == is allowed to call <=>, but this is later disallowed due to efficiency concerns (<=> is generally an inefficient way to implement ==). operator<=>() = default is still defined to implicitly define operator==, which correctly calls == instead of <=> on members, for convenience. So what you want is this:

struct A {
    int n;
    auto operator<=>(const A& rhs) const noexcept = default;
};

// ^^^ basically expands to vvv

struct B {
    int n;
    bool operator==(const B& rhs) const noexcept
    {
        return n == rhs.n;
    }
    auto operator<=>(const B& rhs) const noexcept
    {
        return n <=> rhs.n;
    }
};

请注意，您可以独立地默认 operator == ，同时仍提供用户定义的 operator< => ：

Note that you can independently default operator== while still providing a user-defined operator<=>:

struct B {
    int n;
    // note: return type for defaulted equality comparison operator
    //       must be 'bool', not 'auto'
    bool operator==(const B& rhs) const noexcept = default;
    auto operator<=>(const B& rhs) const noexcept
    {
        return n <=> rhs.n;
    }
};

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