Swift protocol generic as function return type

Question

I want to use generic protocol type as a function return type like this:

protocol P {
  associatedtype T
  func get() -> T?
  func set(v: T)
}

class C<         


        

Answer 1

The problem is you cannot use the syntax P. P is a protocol, meaning it can't be treated as a generic type (Cannot specialize non-generic type 'P'), even though it may have a given associatedtype.

In fact, because it has an associatedtype, you now can't even use the protocol type itself directly – you can only use it as a generic constraint.

One solution to your problem is to simply change your function signature to createC() -> C, as that's exactly what it returns.

class Factory {
    func createC() -> C {
        return C()
    }
}

I'm not entirely sure what you would gain from having the return type be a protocol here. Presumably your example is just a simplification of your actual code and you want to be able to return an arbitrary instance that conforms to P. In that case, you could use type erasure:

class AnyP : P {

    private let _get : () -> T?
    private let _set : (T) -> ()

    init(_ base:U) {
        _get = base.get
        _set = base.set
    }

    func get() -> T? {return _get()}
    func set(v: T) {_set(v)}
}

class Factory {
    func createC() -> AnyP {
        return AnyP(C())
    }
}