How to use three-way comparison (spaceship op) to implement operator== between different types?

Simple Task: I have these two types

struct type_a{
   int member;
};

struct type_b{
   int member;
};

I want to use this new C++20 spaceship op that everyone says is so cool to be able to write type_a{} == type_b{}. I didn’t manage to do that. Even if I write operator<=> between them, I only ever can call type_a{} <=> type_b{}, but never a simple comparison. That confuses me as with a single class, the three-way comparison also defines all the others.

Alternative formulation? How to make it so that std::three_way_comparable_with<type_a, type_b> is true?

>Solution :

The premise of the question is wrong. You don’t use the three-way comparison operator (<=>) to implement ==: you use == to implement ==:

bool operator==(const type_a& a, const type_b& b) {
    return a.member == b.member;
}

The source of confusion is that there is one exception to this rule: if a type declares a defaulted <=> then it also declares a defaulted ==:

struct type_c {
    int member;
    auto operator<=>(type_c const&) const = default;
};

That declaration is equivalent to having written:

struct type_c {
    int member;
    bool operator==(type_c const&) const = default;
    auto operator<=>(type_c const&) const = default;
};

But it’s not the <=> that gives you ==: it’s still the ==, and only ==, that gives you ==.