Issue with Swift 5.7 existentials

Question

I recently converted some of my code which originally used Type-Erasure to use the new Swift 5.7 any existential. However, I'm getting some issues when trying to use the any keyword with already implemented generic types.

I'm running this on Xcode 14 Beta 2 (which has Implicitly Opened Existentials).

Here is an example:

protocol Provider<Value> {
    associatedtype Value
    
    func get() -> Value
}

struct S {
    var stringProvider: any Provider<String>
}

Here is a very simple struct S which has a member stringProvider. I use the any keyword here instead of making any Provider<String> generic because I would like to be able to reassign stringProvider to a different value later on (which has a different type).

struct ProviderView<P: Provider>: View {
    let provider: P
    
    var body: some View {
        Text(String(describing: type(of: provider.get())))
    }
}

Now here I have a ProviderView SwiftUI struct, which takes in a Provider and does stuff with it.

struct DummyProvider: Provider {
    typealias Value = String
    
    func get() -> String {
        "Hello World!"
    }
}

And this is just a dummy Provider implementation which just returns a string.

This all works fine, the problem comes when I try to use ProviderView with an existential any.

struct ContentView: View {
    let s = S(stringProvider: DummyProvider())
    
    var body: some View {
        VStack {
            Image(systemName: "globe")
                .imageScale(.large)
                .foregroundColor(.accentColor)
            
            ProviderView(provider: s.stringProvider) // This is the erroring line
        }
    }
}

I get an error saying Type 'any Provider<String>' cannot conform to 'Provider'.

I think I know why, it's because ProviderView cannot have an any existential as a generic argument.

My question is: Is there any way around this, without going back to type erasure? Am I doing something really badly wrong? Keep in mind that I need to be able to reassign S.stringProvider to a Provider of a different type.

We've got explicit reply from Apple engineer that `any` (ie. anything result in `any View`) does not work for SwiftUI "mechanics". See this thread https://twitter.com/asperi_se/status/1543294437979217920?s=20&t=gIAPoC6fHj_l0P8F_ETLWA — Asperi, Jul 05 '22 at 03:33

score 1 · Accepted Answer · answered Jul 04 '22 at 21:42

The problem here is that body needs to be some specific type, so ProviderView needs to be some specific type (since body relies on that type). It can't change at runtime. So ProviderView can't be generic over its Provider if that's going to change at runtime.

That means that ProviderView.provider needs to be any Provider, not generic:

struct ProviderView: View {
    let provider: any Provider

    var body: some View {
        Text(String(describing: type(of: provider.get())))
    }
}

So that part is expected.

The problem you'll run into is that the current runtime can't quite handle this yet, and you'll get an error later:

// Runtime support for parameterized protocol types is only available in iOS 99.0.0 or newer
ProviderView(provider: s.stringProvider)

I do expect this to improve in later betas, though I encourage you to open a Feedback to track it.

Ok, this makes so much more sense to me now. Thank you so much! I think I'll stick with type erasers though. — , Jul 04 '22 at 21:48

Cristik · Answer 2 · 2022-07-26T12:08:40.083

any Provider is an existential, and existentials being some kind of boxes, don't conform to the protocol inside the box, thus you see the error.

If you're not keen on having ProviderView generic, you can add a custom initializer that uses some - this represents the actual value conforming to the protocol, so the compiler is happy.

struct ProviderView: View {
    let provider: any Provider
    
    init(provider: some Provider) {
        self.provider = provider
    }
    
    var body: some View {
        Text(String(describing: type(of: provider.get())))
    }
}

Though with a generic view like this, there's not much you can accomplish with the provider.

Issue with Swift 5.7 existentials

2 Answers2

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