[swift-evolution] [Proposal] Improving operator requirements in protocols

Pyry Jahkola pyry.jahkola at iki.fi
Tue May 3 01:58:02 CDT 2016

FWIW, the problems with class inheritance and Equatable can be fixed by changing the definition of Equatable into using an associatedtype for the RHS:

protocol Equatable {
    typealias EqualSelf = Self
    func == (lhs: Self, rhs: EqualSelf) -> Bool
    func != (lhs: Self, rhs: EqualSelf) -> Bool
func != <T : Equatable>(lhs: T, rhs: T.EqualSelf) -> Bool {
    return !(lhs == rhs)

This way, a generic function of type Subclass will expect a comparison of type `(Subclass, Subclass.EqualSelf) -> Bool`, i.e. `(Subclass, Base) -> Bool`, so it'll do the right comparison.

I'm not proposing this change because I haven't found but contrived use for it. But if there's demand (and a real use case) for e.g. making an existential of Equatable (say, a struct AnyEquatable), then this could be the fix.

Value types will obviously keep working just like before, since T and T.EqualSelf are synonymous for them.

— Pyry

>> We do have some problems today, such as above where using `==` on a `(Base, Subclass as Base)` pair ends up calling `==(Base, Base)` because we lack multiple dispatch. What surprised me though was that the `eq` call between two `Subclass` instances passed to the trampoline operator ended up calling `Base.eq`. I would have expected `Subclass.eq` to be called there since the generic argument `T` was bound to `Subclass`. Today, a non-generic `==(Subclass, Subclass)` operator *does* do the right thing.
> The reason why the trampoline calls the ==(Base,Base) overload in
> Subclass is because it is a part of the override chain for the base
> class method that introduces the conformance.
> The reason why the non-generic ==(Subclass,Subclass) operator does the
> right thing is because it is a better match for overload resolution of
> == at the callsite.  But if you have a (Subclass as Base, Subclass as
> Base) pair, even today it will call ==(Base,Base).

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