ios – AppIntents problem when importing UIKit and utilizing an Picture for caseDisplayRepresentations of AppEnum

I’m making a easy AppIntents Shortcut that requires UIKit to open a URL in my app. Although, as soon as I add the import UIKit to my file, Xcode exhibits me the next warning on the caseDisplayRepresentations, as a result of I take advantage of a picture for the show repreentations. With out importing UIKit (or with out the picture for the DisplayRepresentations) all the things works effective! Any concepts learn how to repair that?

Anticipated a direct name to the Picture initializer, obtained Swift.Optionally available<AppIntents.DisplayRepresentation.Picture> as a substitute

The Code

import AppIntents
import UIKit

enum AppView: Int, AppEnum {
    case tab1
    case tab2
    
    static var typeDisplayRepresentation = TypeDisplayRepresentation(identify: "View")
    
    static var caseDisplayRepresentations: [AppView : DisplayRepresentation] = [
        .tab1: DisplayRepresentation(title: "Tab 1",
                                     image: .init(systemName: "1.circle")),
        .tab2: DisplayRepresentation(title: "Tab 2",
                                     image: .init(systemName: "2.circle"))
    ]
}

// MARK: - Intent
struct OpenAppIntent: AppIntent {
    static let title: LocalizedStringResource = "Open App"
    static let description: IntentDescription = "Opens the app within the given view."
    
    // Launches app when motion is triggered
    static let openAppWhenRun: Bool = true
    
    // App View Parameter
    @Parameter(title: "View",
               description: "The view contained in the app.",
               default: .tab1,
               requestValueDialog: "The place would you wish to navigate to?")
    var view: AppView
    
    // Shortcuts Motion Textual content
    static var parameterSummary: some ParameterSummary {
        Abstract("Open (.$view)")
    }
    
    
    @MainActor
    func carry out() async throws -> some IntentResult {
        if let url = URL(string: "myapp://(view)") {
            await UIApplication.shared.open(url)
        }
        
        return .end result()
    }
}