Having problems trying out the example code in this session

Question

real_kaipi OP

Created Jun ’22

Replies 9

Boosts 1

Views 2.5k

Participants 7

I've copied the code as the video was coming along, and getting this error:

 error: type 'any Animal' cannot conform to 'Animal'
note: only concrete types such as structs, enums and classes can conform to protocols
note: required by instance method 'feed' where 'some Animal' = 'any Animal'

is this functionality not yet available in Xcode 14b1?

Boost

Answer 1

Claude31 OP

Jun ’22

Which code ? There are billions on the internet. If you don't show, how could we guess ?

You are apparently new to the forum, so read this: https://developer.apple.com/forums/thread/706527

0

Answer 2

real_kaipi OP

Jun ’22

 import Cocoa
 
protocol AnimalFeed {
 static func grow() -> Self
}
 
protocol Animal {
 associatedtype Feed: AnimalFeed
 
 func eat(_ food: Feed)
}
 
struct Hay: AnimalFeed {
 static func grow() -> Self {
  print("growing hay")
  return Hay()
 }
}
 
struct Pellet: AnimalFeed {
 static func grow() -> Self {
  print("growing pellets")
  return Pellet()
 }
}
 
struct Horse: Animal {
 func eat(_ food: Hay) {
  print("horse eating \(food)")
 }
}
 
struct Dog: Animal {
 func eat(_ food: Pellet) {
  print("dog eating \(food)")
 }
}
 
struct Farm {
 func feed(_ animal: some Animal) {
  let food = type(of: animal).Feed.grow()
  animal.eat(food)
 }
 
 func feedAll(_ animals: [any Animal]) {
  for animal in animals {
   feed(animal)
  }
 }
}
 
let animals: [any Animal] = [
 Horse(),
 Horse(),
 Dog(),
]
 
let farm = Farm()
farm.feedAll(animals)

according to the video, that should compile...

aside: I opened the forum post creator from the Embrace Swift Generics video, which added the tag for the video, so I assumed it would signal which video I was talking about, but I guess the forum is not very useful in that regard

2

Answer 3

Claude31 OP

Jun ’22

Hope I'm not wrong here. And I would be interested by the correct answer to this post.

You are applying any to a protocol. I read that would only be availabale with Swift 6 (Xcode 14 is Swift 5.7 : https://swiftversion.net/).

However, example given in SE-0352 (implemented in Swift 5.7) tells it is possible

https://github.com/apple/swift-evolution/blob/main/proposals/0352-implicit-open-existentials.md

And you're right, it is very similar to SE-0352, so it should work.

Unless it is not yet in 14 beta ?

If that may help, I noted that this works:

     func feedAll(_ animals: [any Animal]) {
        for animal in animals {
            print("all ->", animal) // feed(animal)
            if animal is Horse {
                feed(Horse())
            }
            if animal is Dog {
                feed(Dog())
            }
        }
    }

1

Answer 4

real_kaipi OP

Jun ’22

This is pretty much copy paste (some unrelated minor modifications) from the referenced video, that mentioned this is available in 5.7 - Xcode 14.

0

Answer 5

TomSwift1973 OP

Jun ’22

I'm having the same issue with the same XCode beta. I watched the video then tried with my own example and couldn't make it work. Provided sample doesn't work either.

0

Answer 6

Developer Tools Engineer OP

Apple

Jun ’22

Thank you for reporting this bug! Going into your Xcode build settings, filtering by "Other Swift Flags", and adding -Xfrontend -enable-experimental-opened-existential-types will work around the issue.

1

Answer 7

Developer Tools Engineer OP

Apple

Jun ’22

I'm unable to reproduce the issue by copy-pasting the complete example code from Embrace Swift Generics at 27:10 into a new command-line macOS tool in Xcode 14 Beta 1. Could you please file a Feedback report, attaching the project and the failed build log, so I can investigate this further? Thank you!

0

Answer 8

zunda OP

Jun ’22

I face to same issue in Playground. Normal Project can compile it.

1

Answer 9

DalekSein OP

Jun ’22

I had another situation involving SwiftUI with the same error message: Type 'any Identifiable' cannot conform to 'Identifiable'.

I was building on normal project with Xcode 14 beta 2. The project doesn't build with or without the compiler flag -Xfrontend -enable-experimental-opened-existential-types as suggested in prev comment.

 import SwiftUI
 
struct A: Identifiable{
    let id = UUID()
}
 
struct B: Identifiable {
    let id = UUID()
}
 
struct ContentView: View {
    let container: [any Identifiable] = [A(), B()]
    var body: some View {
        VStack {
            ForEach(container) { item in
                Text("\(item.id)")
            }
        }
    }
}

0

	error: type 'any Animal' cannot conform to 'Animal'
	note: only concrete types such as structs, enums and classes can conform to protocols
	note: required by instance method 'feed' where 'some Animal' = 'any Animal'

	import Cocoa

	protocol AnimalFeed {
	static func grow() -> Self
	}

	protocol Animal {
	associatedtype Feed: AnimalFeed

	func eat(_ food: Feed)
	}

	struct Hay: AnimalFeed {
	static func grow() -> Self {
	print("growing hay")
	return Hay()
	}
	}

	struct Pellet: AnimalFeed {
	static func grow() -> Self {
	print("growing pellets")
	return Pellet()
	}
	}

	struct Horse: Animal {
	func eat(_ food: Hay) {
	print("horse eating \(food)")
	}
	}

	struct Dog: Animal {
	func eat(_ food: Pellet) {
	print("dog eating \(food)")
	}
	}

	struct Farm {
	func feed(_ animal: some Animal) {
	let food = type(of: animal).Feed.grow()
	animal.eat(food)
	}

	func feedAll(_ animals: [any Animal]) {
	for animal in animals {
	feed(animal)
	}
	}
	}

	let animals: [any Animal] = [
	Horse(),
	Horse(),
	Dog(),
	]

	let farm = Farm()
	farm.feedAll(animals)

	func feedAll(_ animals: [any Animal]) {
	for animal in animals {
	print("all ->", animal) // feed(animal)
	if animal is Horse {
	feed(Horse())
	}
	if animal is Dog {
	feed(Dog())
	}
	}
	}

	import SwiftUI

	struct A: Identifiable{
	let id = UUID()
	}

	struct B: Identifiable {
	let id = UUID()
	}

	struct ContentView: View {
	let container: [any Identifiable] = [A(), B()]
	var body: some View {
	VStack {
	ForEach(container) { item in
	Text("\(item.id)")
	}
	}
	}
	}