'How to make a SwiftUI View a property of a Protocol
I'm trying to extend a protocol so that a certain few impls of the protocol have a view associated with them. However, because a SwiftUI View is a protocol, this is proving to be a challenge.
import SwiftUI
protocol ParentProtocol {
var anyProperty: String { get }
}
protocol ChildProtocol : ParentProtocol {
associatedtype V
var someView: V { get }
}
class ChildImpl : ChildProtocol {
var someView : some View {
Text("Hello World")
}
var anyProperty: String = ""
}
class ChildMgr {
var child: ParentProtocol = ChildImpl()
func getView() -> some View {
guard let child = child as? ChildProtocol else { return EmptyView() }
return child.someView
}
}
Its not clear to me where to constrain the ChildProtocol's associated type to a View (or Text for that matter).
At the guard let child = ... I get the following compiler error:
Protocol 'ChildProtocol' can only be used as a generic constraint because it has Self or associated type requirements
and when returning the chid's view I get:
Member 'someView' cannot be used on value of protocol type 'ChildProtocol'; use a generic constraint instead
I think the answer may be in this thread: https://developer.apple.com/forums/thread/7350 but frankly its confusing on how to apply it to this situation.
Solution 1:[1]
Don't use runtime checks. Use constrained extensions.
I also don't see a reason for you to be using classes.
protocol ChildProtocol: ParentProtocol {
associatedtype View: SwiftUI.View
var someView: View { get }
}
final class ChildImpl: ChildProtocol {
var someView: some View {
Text("Hello World")
}
var anyProperty: String = ""
}
final class ChildMgr<Child: ParentProtocol> {
var child: Child
init(child: Child) {
self.child = child
}
}
extension ChildMgr where Child: ChildProtocol {
func getView() -> some View {
child.someView
}
}
extension ChildMgr {
func getView() -> some View {
EmptyView()
}
}
extension ChildMgr where Child == ChildImpl {
convenience init() {
self.init(child: .init())
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Jessy |
