CODEWITHSUNDEEP
swiftgenericsprotocols
Valentyn Zakharenko
Valentyn Zakharenko

Reputation: 3088

How to implement a generic class that conforms one of the mutually associated protocols?

I have two protocols. Each (x) of them contains associatedtype that requires to confirm another protocol (y) when associatedtype of this protocol (y) equals Self (of x).

protocol B {
    associatedtype AA: A
        where AA.BB == Self
}

protocol A {
    associatedtype BB: B
        where BB.AA == Self
}

It is no problem to implement those protocols as follows:

class AAA: A {
    typealias BB = BBB
}

class BBB: B {
    typealias AA = AAA
}

But I can't implement one of them as generic.

class AAA<BBBB: B>: A {
    typealias BB = BBBB
}

class BBB: B {
    typealias AA = AAA<BBB>
}

As a result, I have an error:

'A' requires the types 'AAA' and 'BBBB.AA' be equivalent

And note:

requirement specified as 'Self' == 'Self.BB.AA' [with Self = AAA]

Sounds understandable. I added the constraint to my generic class.

class AAA<BBBB: B>: A where BBBB.AA == AAA {
    typealias BB = BBBB
}

And result

error: type 'AAA' does not conform to protocol 'A'
note: protocol requires nested type 'BB'; do you want to add it?

error: type 'BBB' does not conform to protocol 'B'
note: protocol requires nested type 'AA'; do you want to add it?

All my tries to solve it was unsuccessful, they just changed error messages.
Is it possible to do? How?

Upvotes: 1

Views: 86

Answers (1)

damdamo
damdamo

Reputation: 353

With my current knowledge, I don't think it's possible to do what you want. I was looking for more information and I found some of them about why it's not needed.

Why Swift doesn't have F-Bounded.

Support for F-bounded polymorphism?

Upvotes: 1

Related Questions