Unbalanced calls to begin/end appearance transitions for

Question

I read SO about another user encountering similar error, but this error is in different case.

I received this message when I added a View Controller initially:

Answer 1

If you're using transitioningDelegate (not the case in this question's example), also set modalPresentationStyle to .Custom.

Swift

let vc = storyboard.instantiateViewControllerWithIdentifier("...")
vc.transitioningDelegate = self
vc.modalPresentationStyle = .Custom

Answer 2

I had lot of problem with the same issue. I solved this one by

1. Initiating the ViewController using the storyboad instantiateViewControllerWithIdentifier method. i.e Intro *vc = [self.storyboard instantiateViewControllerWithIdentifier:@"introVC"];
2. [self.tabBarController presentModalViewController : vc animated:YES];

I have the viewcontroller in my storyboard, for some reason using only [[introvc alloc] init]; did not work for me.

Answer 3

I had the same problem. I called a method inside viewDidLoad inside my first UIViewController

- (void)viewDidLoad{
    [super viewDidLoad];

    [self performSelector:@selector(loadingView)
               withObject:nil afterDelay:0.5];
}

- (void)loadingView{

    [self performSegueWithIdentifier:@"loadedData" sender:self];
}

Inside the second UIViewController I did the same also with 0.5 seconds delay. After changing the delay to a higher value, it worked fine. It's like the segue can't be performed too fast after another segue.

Answer 4

I had the same problem when I need to Present My Login View Controller from another View Controller If the the User is't authorized, I did it in ViewDidLoad Method of my Another View Controller ( if not authorized -> presentModalViewController ). When I start to make it in ViewDidAppear method, I solved this problem. I Think that ViewDidLoad only initialize properties and after that the actual showing view algorithm begins! Thats why you must use viewDidAppear method to make modal transitions!

Answer 5

As @danh suggested, my issue was that I was presenting the modal vc before the UITabBarController was ready. However, I felt uncomfortable relying on a fixed delay before presenting the view controller (from my testing, I needed to use a 0.05-0.1s delay in performSelector:withDelay:). My solution is to add a block that gets called on UITabBarController's viewDidAppear: method:

PRTabBarController.h:

@interface PRTabBarController : UITabBarController

@property (nonatomic, copy) void (^viewDidAppearBlock)(BOOL animated);

@end

PRTabBarController.m:

#import "PRTabBarController.h"

@implementation PRTabBarController

- (void)viewDidAppear:(BOOL)animated {
    [super viewDidAppear:animated];
    if (self.viewDidAppearBlock) {
        self.viewDidAppearBlock(animated);
    }
}

@end

Now in application:didFinishLaunchingWithOptions:

PRTabBarController *tabBarController = [[PRTabBarController alloc] init];

// UIWindow initialization, etc.

__weak typeof(tabBarController) weakTabBarController = tabBarController;
tabBarController.viewDidAppearBlock = ^(BOOL animated) {
    MyViewController *viewController = [MyViewController new];
    viewController.modalPresentationStyle = UIModalPresentationOverFullScreen;
    UINavigationController *navigationController = [[UINavigationController alloc] initWithRootViewController:viewController];
    [weakTabBarController.tabBarController presentViewController:navigationController animated:NO completion:nil];
    weakTabBarController.viewDidAppearBlock = nil;
};

Answer 6

As posted by danh

You can generate this warning by presenting the modal vc before the app is done initializing. i.e. Start a tabbed application template app and present a modal vc on top of self.tabBarController as the last line in application:didFinishLaunching. Warning appears. Solution: let the stack unwind first, present the modal vc in another method, invoked with a performSelector withDelay:0.0

Try to move the method into the viewWillAppear and guard it so it does get executed just once (would recommend setting up a property)