为什么我可以用默认的而不能用用户提供的调用==？

Question

#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
}

用clang-10编译为clang-std=c++20-stdlib=libc++++main.cpp

Answer 1

在宇宙飞船操作员的最初设计中，允许==调用<=>，但后来出于效率考虑不允许调用(<=>通常是实现==)的低效方法。操作符<=>()=default仍然被定义为隐式定义操作符==，为了方便起见，它正确地调用成员的==而不是<=>。所以你想要的是：

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==，同时仍然提供用户定义的操作符<=>:

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;
    }
};