Protocols inheritance and typing incomprehension

Question

I am a bit confused. Please look at this example.

I created a VM protocol:

protocol VM {

}

And this protocol is using in my VC implementation

final class VC: UIViewController {
    let viewModel: VM
}

Now I create special new protocols

protocol AwesomeProtocol {

}

protocol AwesomeViewProtocol {
     var viewModel: AwesomeProtocol { get }
}

My idea is to expand VM with Awesomeness so:

protocol VM: AwesomeProtocol {

}

final class VC: UIViewController, AwesomeViewProtocol {
    let viewModel: VM
}

But here I met an compiler error:

Type 'VC' does not conform to protocol 'AwesomeViewProtocol'

Despite the fact that VM extend AwesomeProtocol

Someone could explain me what am I doing wrong?

Compare https://stackoverflow.com/q/42561685/2976878 – Hamish May 16 '18 at 10:14 — Hamish, May 16 '18 at 10:14

score 1 · Accepted Answer · answered May 16 '18 at 10:09

1

You have to implement this.

final class VC: UIViewController, AwesomeViewProtocol {
    var viewModel: AwesomeProtocol
}

computed variables are close to the functions. Their signatures must be the same in a parent and child (inherited) classes/protocols.

If you need something abstracts use assosiatedtype and generic classes instead.

answered May 16 '18 at 10:09

Vyacheslav

In your implementation `VC` lose access to `VM` methods. It will can use only `AwesomeProtocol` interface. According to your answer I assume that my implementation idea is not possible and the only case is to use mentioned `assosiatedtype`. Am I right? – Kamil Harasimowicz May 16 '18 at 10:15
@KamilHarasimowicz I think, yes – Vyacheslav May 16 '18 at 10:16

1 Answers1