Synthesis of == from operator<=>

Asperamanca

Consider this class:

class X
{
public:
    int m_X = 0;
    int m_Y = 0;
    
    std::strong_ordering operator<=>(const X& other) const = default;
};

I can now write

X a;
X b;
auto equal = (a == b);

Perfect.
However, when I have to implement operator<=> myself, this no longer works:

class X
{
public:
    int m_X = 0;
    int m_Y = 0;
    
    std::strong_ordering operator<=>(const X& other) const
    {
        return std::tie(m_X, m_Y) <=> std::tie(other.m_X, other.m_Y);
    }
};

I get compiler errors such as
error: C2676: binary '==': 'X' does not define this operator or a conversion to a type acceptable to the predefined operator
(MSVC 2022), or
error: no match for 'operator==' (operand types are 'X' and 'X')
(MinGW 11.2)

I have been digging around cppreference a bit, and also took a look at P0515 but couldn't come up with a clear idea why one works and the other does not.
Can someone please explain?

JoeCFD

@Asperamanca from here https://cplusplus.com/doc/tutorial/operators/
<=> is available with C++20

Are you building it on VS or MinGW?

SGaist

Hi,

It's by design. The TLDR: if you declare a custom spaceship operator, you have to declare the equality operator. If that operator should use the default implementation then just declare it as default.

See this StackOverflow thread for more details.

Asperamanca

@SGaist Many thanks, I must have missed that one!