How to write custom spaceship operator in c++20 [duplicate] - c++

Question

I'm running into a strange behavior with the new spaceship operator <=> in C++20. I'm using Visual Studio 2019 compiler with /std:c++latest.
This code compiles fine, as expected:
#include <compare>
struct X
{
int Dummy = 0;
auto operator<=>(const X&) const = default; // Default implementation
};
int main()
{
X a, b;
a == b; // OK!
return 0;
}
However, if I change X to this:
struct X
{
int Dummy = 0;
auto operator<=>(const X& other) const
{
return Dummy <=> other.Dummy;
}
};
I get the following compiler error:
error C2676: binary '==': 'X' does not define this operator or a conversion to a type acceptable to the predefined operator
I tried this on clang as well, and I get similar behavior.
I would appreciate some explanation on why the default implementation generates operator== correctly, but the custom one doesn't.

Answer 1

This is by design.
[class.compare.default] (emphasis mine)
3 If the class definition does not explicitly declare an ==
operator function, but declares a defaulted three-way comparison
operator function, an == operator function is declared implicitly
with the same access as the three-way comparison operator function.
The implicitly-declared == operator for a class X is an inline
member and is defined as defaulted in the definition of X.
Only a defaulted <=> allows a synthesized == to exist. The rationale is that classes like std::vector should not use a non-defaulted <=> for equality tests. Using <=> for == is not the most efficient way to compare vectors. <=> must give the exact ordering, whereas == may bail early by comparing sizes first.
If a class does something special in its three-way comparison, it will likely need to do something special in its ==. Thus, instead of generating a potentially non-sensible default, the language leaves it up to the programmer.

Answer 2

During the standardization of this feature, it was decided that equality and ordering should logically be separated. As such, uses of equality testing (== and !=) will never invoke operator<=>. However, it was still seen as useful to be able to default both of them with a single declaration. So if you default operator<=>, it was decided that you also meant to default operator== (unless you define it later or had defined it earlier).
As to why this decision was made, the basic reasoning goes like this. Consider std::string. Ordering of two strings is lexicographical; each character has its integer value compared against each character in the other string. The first inequality results in the result of ordering.
However, equality testing of strings has a short-circuit. If the two strings aren't of equal length, then there's no point in doing character-wise comparison at all; they aren't equal. So if someone is doing equality testing, you don't want to do it long-form if you can short-circuit it.
It turns out that many types that need a user-defined ordering will also offer some short-circuit mechanism for equality testing. To prevent people from implementing only operator<=> and throwing away potential performance, we effectively force everyone to do both.

Answer 3

The other answers explain really well why the language is like this. I just wanted to add that in case it's not obvious, it is of course possible to have a user-provided operator<=> with a defaulted operator==. You just need to explicitly write the defaulted operator==:
struct X
{
int Dummy = 0;
auto operator<=>(const X& other) const
{
return Dummy <=> other.Dummy;
}
bool operator==(const X& other) const = default;
};
Note that the defaulted operator== performs memberwise == comparisons. That is to say, it is not implemented in terms of the user-provided operator<=>. So requiring the programmer to explicitly ask for this is a minor safety feature to help prevent surprises.