ios - 取消向后滑动时，SwiftUI 应用程序卡住，100% CPU

Question

以下代码可能会使正在运行的应用程序在 iPhone(而非 iPad)或 iPhone 模拟器上无响应。 Xcode 显示该应用程序消耗 100% CPU，同时分配越来越多的内存。

struct SecondView: View {
    @State private var keyboardHeight: CGFloat = 0

    private let showPublisher = NotificationCenter.Publisher.init(
        center: .default,
        name: UIResponder.keyboardWillShowNotification
    ).map { (notification) -> CGFloat in
        if let rect = notification.userInfo?["UIKeyboardFrameEndUserInfoKey"] as? CGRect {
            return rect.size.height
        } else {
            return 0
        }
    }

    var body: some View {
        VStack(spacing: 20) {
            if keyboardHeight == 0 {
                Text("This is shown as long as there's no keyboard")
            }
            Text("This is the SecondView. Drag from the left edge to navigate back, but don't complete the gesture: crash results.")
        }.onReceive(self.showPublisher) { (height) in
            self.keyboardHeight = height
        }
        .navigationBarItems(trailing: Button("Dummy") {  })
    }
}

struct ContentView: View {
    @State var textInput = ""

    var body: some View {
        NavigationView {
            VStack(spacing: 20) {
                TextField("1. Tap here to show keyboard", text: self.$textInput)
                    .textFieldStyle(RoundedBorderTextFieldStyle())
                NavigationLink(destination: SecondView()) {
                    Text("2. Go to second screen")
                }
                Spacer()
            }
        }
    }
}

触发卡住:

点击文本框使键盘出现

点击链接转到下一个屏幕

从屏幕左侧拖动，但不要完成手势，而是提前释放

有一些解决方法:

删除 SecondView 中的导航栏项目(虚拟按钮)

删除使用keyboardHeight SecondView 中的变量

不要在ContentView中激活键盘导航前

但是，我无法在我的应用程序中使用上述解决方法。有谁知道根本原因是什么？

Answer 1

我能够在导航之前停用键盘，使用以下解决方法:

NavigationLink(destination: SecondView(), tag: 2, selection: $navigationSelection) {
    EmptyView()
}
Text("2. Go to second screen")
    .onTapGesture {
        UIApplication.shared.sendAction(#selector(UIResponder.resignFirstResponder), to: nil, from: nil, for: nil)
        self.navigationSelection = 2
    }