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iOS: How do I know if a property is KVO-compliant?

In the Key-Value Observing Programming Guide, the section Registering for Key-Value Observing says "Typically properties in Apple-supplied frameworks are only KVO-compliant if they are documented as such." But, I haven't found any properties in the documentation that are documented as KVO-compliant. Would you please point me to some?

Specifically, I would like to know if the @property(nonatomic,retain) UIViewController *rootViewController of UIWindow is KVO-compliant. The reason is that I'm adding the rootViewController property to UIWindow for iOS < 4 and want to know if I should make it KVO-compliant.

@interface UIWindow (Additions)

#if __IPHONE_OS_VERSION_MIN_REQUIRED < __IPHONE_4_0
@property (nonatomic, retain) UIViewController *rootViewController;
#endif;

@end

@implementation UIWindow (Additions)

#if __IPHONE_OS_VERSION_MIN_REQUIRED < __IPHONE_4_0
@dynamic rootViewController;

- (void)setRootViewController:(UIViewController *)newRootViewController {
    if (newRootViewController != _rootViewController) {
        // Remove old views before adding the new one.
        for (UIView *subview in [self subviews]) {
            [subview removeFromSuperview];
        }
        [_rootViewController release];
        _rootViewController = newRootViewController;
        [_rootViewController retain];
        [self addSubview:_rootViewController.view];
    }
}
#endif

@end
like image 248
ma11hew28 Avatar asked Jul 07 '11 14:07

ma11hew28


1 Answers

Short answer: No.

Long answer: Nothing in UIKit is guaranteed to be KVO-compliant. If you happen to find that KVO-ing a property works, be grateful, it's unintentional. Also: be wary. It could very well break in the future.

If you find that this is something you need, please file an enhancement request.


About your actual code, it's inherently flawed. Do NOT attempt to add a "rootViewController" setter to UIWindow this way. It will break when you compile your code on iOS 4 but someone runs it on an iOS 5 device. Because you compiled using the 4.x SDK, the #if statements will evaluate to true, meaning your category method smasher will be included in the binary. However, when you run it on an iOS 5 device, you're now going to get a method conflict because two methods on UIWindow will have the same method signature, and there's no guarantee as to which one will be used.

Don't screw with the frameworks like this. If you have to have this, use a subclass. THIS IS WHY SUBCLASSING EXISTS.


Your subclass would look something like this:

@interface CustomWindow : UIWindow

@property (nonatomic, retain) UIViewController *rootViewController;

@end

@implementation CustomWindow : UIWindow

static BOOL UIWindowHasRootViewController = NO;

@dynamic rootViewController;

- (void)_findRootViewControllerMethod {
  static dispatch_once_t predicate;
  dispatch_once(&predicate, ^{
    IMP uiwindowMethod = [UIWindow instanceMethodForSelector:@selector(setRootViewController:)];
    IMP customWindowMethod = [CustomWindow instanceMethodForSelector:@selector(setRootViewController:)];
    UIWindowHasRootViewController = (uiwindowMethod != NULL && uiwindowMethod != customWindowMethod);
  });
}

- (UIViewController *)rootViewController {
  [self _findRootViewControllerMethod];
  if (UIWindowHasRootViewController) {
    // this will be a compile error unless you forward declare the property
    // i'll leave as an exercise to the reader ;)
    return [super rootViewController];
  }
  // return the one here on your subclass
}

- (void)setRootViewController:(UIViewController *)rootViewController {
  [self _findRootViewControllerMethod];
  if (UIWindowHasRootViewController) {
    // this will be a compile error unless you forward declare the property
    // i'll leave as an exercise to the reader ;)
    [super setRootViewController:rootViewController];
  } else {
    // set the one here on your subclass
  }
}

Caveat Implementor: I typed this in a browser window

like image 107
Dave DeLong Avatar answered Oct 10 '22 22:10

Dave DeLong