为什么默认三通运算符（spaceship <=>）生成相等运算符（==）而用户定义三通运算符不？

Question

Consider this code:

#include <iostream>
#include <compare>

class A {
public:
  int i = {};

  std::strong_ordering operator<=> (A const& r) const
  {
    return i <=> r.i;
  }
};

void TestA()
{
    A a;
    A b;

    std::cout<< (a<b);    
    std::cout<< (a>b);
    std::cout<< (a<=b);
    std::cout<< (a>=b);
    //std::cout<< (a==b); //ERROR
    std::cout << 'E';
    //std::cout<< (a!=b); //ERROR
    std::cout << 'E';
    std::cout<< std::is_eq(a<=>b);
    std::cout<< std::is_neq(a<=>b) << std::endl;
}

class B {
public:
  int i = {};

  std::strong_ordering operator<=> (B const& r) const = default;

};


void TestB()
{
    B a;
    B b;

    std::cout<< (a<b);    
    std::cout<< (a>b);
    std::cout<< (a<=b);
    std::cout<< (a>=b);
    std::cout<< (a==b);
    std::cout<< (a!=b);
    std::cout<< std::is_eq(a<=>b);
    std::cout<< std::is_neq(a<=>b) << std::endl;
}

class C {
public:
  bool b = {};
  int v1 = {};
  int v2 = {};

  std::strong_ordering operator<=> (C const& r) const
  {
      return (b?v1:v2) <=> (r.b?r.v1:r.v2);
  }

  bool operator== (C const& r) const
  {
      return std::is_eq(*this<=>r);
  }

};

void TestC()
{
    C a;
    C b;

    std::cout<< (a<b);    
    std::cout<< (a>b);
    std::cout<< (a<=b);
    std::cout<< (a>=b);
    std::cout<< (a==b);
    std::cout<< (a!=b);
    std::cout<< std::is_eq(a<=>b);
    std::cout<< std::is_neq(a<=>b) << std::endl;
}


int main()
{    
    TestA();
    TestB();
    TestC();

    return 0;
}

https://wandbox.org/permlink/SLmLZOc18RaJV7Mu

Remove comments to get an error.

First I want to ask why default three-way operator behaves differently than user define operator?

And second, is the solution to this problem correct for class C or should it be handled differently?

This is just a simple example and I have more complex situation in mind with tens of fields and unions (If you don't know what I mean, check out some Intel APIs;) ).

Edit:

This question Equality operator does not get defined for a custom spaceship operator implementation in C++20 focused on why there is no default equality operator for user defined 3-way operator, I would like to know why there is a difference in default and user define behavior?

Edit 2:

I slightly modified class C in the example to picture more of real life problem (when default operators are not valid solution). I also want to clarify that I would like to known the reasons behind those differences (between user define and default operator) to be able to assess if my real life solution is correct (similar to C ) as I do value more code maintainability than performance for a part of code that I am working right now.

Answer 1

The principle reason why equality and ordering are separated is performance. If you have a type whose ordering operations are user-defined, then more often than not, you can write a user-defined equality test operation that is more efficient at doing equality tests. And therefore, the language should encourage you to write it by not using operator<=> for direct equality testing.

This only really applies to user-defined ordering/equality operations. Default ordering is member-wise, and default equality operations are also member-wise. And since ordering implies equality, it is reasonable that defaulting ordering also defaults equality.

Yes, they could make people spell it out, but there wasn't really a good reason for that . operator<=> was meant to make it easy to opt-in to default ordering; making you write two declarations for something that's already implied by one of them doesn't make sense.