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 stackoverflow.com/q/42561685/2976878

Hamish
– Hamish

2018-05-16 10:14:40 +00:00
Commented May 16, 2018 at 10:14 — Hamish
– Hamish, Commented May 16, 2018 at 10:14

Vyacheslav · Accepted Answer · 2018-05-16 10:09:37Z

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, 2018 at 10:09

Vyacheslav

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2 Comments

Kamil Harasimowicz Over a year ago

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?

Vyacheslav Over a year ago

@KamilHarasimowicz I think, yes

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