你可以调用你的segue或present, push代码在这个块中:

override func viewDidLoad() {
    super.viewDidLoad()
    OperationQueue.main.addOperation {
        // push or present the page inside this block
    }
}

In case it helps anyone, my issue was extremely silly. Totally my fault of course. A notification was triggering a method that was calling the modal. But I wasn't removing the notification correctly, so at some point, I would have more than one notification, so the modal would get called multiple times. Of course, after you call the modal once, the viewcontroller that calls it it's not longer in the view hierarchy, that's why we see this issue. My situation caused a bunch of other issue too, as you would expect.

总之，无论你做什么，都要确保模态不会被调用超过一次。

我也有这个问题，但这和时间无关。我使用一个单例来处理场景，并将其设置为presenter。换句话说，“自我”没有连接到任何东西。我只是把它的内部“场景”变成了新的主持人，瞧，它成功了。(瞧，当你知道它的意思后，它就失去了它的触感，嘿)。

所以，这不是关于“神奇地找到正确的方法”，而是关于理解你的代码所处的位置以及它在做什么。我很高兴苹果给出了如此直白的警告信息，甚至还带有感情色彩。向苹果开发者致敬!!

我的问题是，在我调用窗口上的makeKeyAndVisible()之前，我在UIApplicationDelegate的didFinishLaunchingWithOptions方法中执行segue。

在主窗口中，可能总是会出现与显示警报不兼容的过渡。为了允许在应用程序生命周期的任何时候显示警报，您应该有一个单独的窗口来完成这项工作。

/// independant window for alerts
@interface AlertWindow: UIWindow

+ (void)presentAlertWithTitle:(NSString *)title message:(NSString *)message;

@end

@implementation AlertWindow

+ (AlertWindow *)sharedInstance
{
    static AlertWindow *sharedInstance;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        sharedInstance = [[AlertWindow alloc] initWithFrame:UIScreen.mainScreen.bounds];
    });
    return sharedInstance;
}

+ (void)presentAlertWithTitle:(NSString *)title message:(NSString *)message
{
    // Using a separate window to solve "Warning: Attempt to present <UIAlertController> on <UIViewController> whose view is not in the window hierarchy!"
    UIWindow *shared = AlertWindow.sharedInstance;
    shared.userInteractionEnabled = YES;
    UIViewController *root = shared.rootViewController;
    UIAlertController *alert = [UIAlertController alertControllerWithTitle:title message:message preferredStyle:UIAlertControllerStyleAlert];
    alert.modalInPopover = true;
    [alert addAction:[UIAlertAction actionWithTitle:@"OK" style:UIAlertActionStyleCancel handler:^(UIAlertAction *action) {
        shared.userInteractionEnabled = NO;
        [root dismissViewControllerAnimated:YES completion:nil];
    }]];
    [root presentViewController:alert animated:YES completion:nil];
}

- (instancetype)initWithFrame:(CGRect)frame
{
    self = [super initWithFrame:frame];

    self.userInteractionEnabled = NO;
    self.windowLevel = CGFLOAT_MAX;
    self.backgroundColor = UIColor.clearColor;
    self.hidden = NO;
    self.rootViewController = UIViewController.new;

    [NSNotificationCenter.defaultCenter addObserver:self
                                           selector:@selector(bringWindowToTop:)
                                               name:UIWindowDidBecomeVisibleNotification
                                             object:nil];

    return self;
}

/// Bring AlertWindow to top when another window is being shown.
- (void)bringWindowToTop:(NSNotification *)notification {
    if (![notification.object isKindOfClass:[AlertWindow class]]) {
        self.hidden = YES;
        self.hidden = NO;
    }
}

@end

在设计上总是成功的基本用法:

[AlertWindow presentAlertWithTitle:@"My title" message:@"My message"];

在我的情况下，我不能把我的类重写。所以，这是我得到的:

let viewController = self // I had viewController passed in as a function,
                          // but otherwise you can do this


// Present the view controller
let currentViewController = UIApplication.shared.keyWindow?.rootViewController
currentViewController?.dismiss(animated: true, completion: nil)

if viewController.presentedViewController == nil {
    currentViewController?.present(alert, animated: true, completion: nil)
} else {
    viewController.present(alert, animated: true, completion: nil)
}