Question

我正在关注https://github.com/RabbitMC/inbox-replica以制作收件箱副本转换的Objective-C版本，但是我陷入了无法制作Objective-C代码的代码。

这是Swift代码：

func animateTransition(transitionContext: UIViewControllerContextTransitioning) {
    let duration = transitionDuration(transitionContext)
    let fromViewController = transitionContext.viewControllerForKey(UITransitionContextFromViewControllerKey)!
    let toViewController = transitionContext.viewControllerForKey(UITransitionContextToViewControllerKey)!
    let containerView = transitionContext.containerView()


    var foregroundViewController = toViewController
    var backgroundViewController = fromViewController

    if type == .Dismissing {
        foregroundViewController = fromViewController
        backgroundViewController = toViewController
    }

    // get target view
    var targetViewController = backgroundViewController
    if let navController = targetViewController as? UINavigationController {
        targetViewController = navController.topViewController!
    }

    let targetViewMaybe = (targetViewController as? ExpandingTransitionPresentingViewController)?.expandingTransitionTargetViewForTransition(self)

这是我的Objective-C代码：

-(void)animateTransition:(id<UIViewControllerContextTransitioning>)transitionContext {
    NSTimeInterval duration = [self transitionDuration:transitionContext];
    UIViewController *fromViewController = [transitionContext viewControllerForKey:UITransitionContextFromViewControllerKey];
    UIViewController *toViewController = [transitionContext viewControllerForKey:UITransitionContextToViewControllerKey];
    UIView *containerView = [transitionContext containerView];

    UIViewController *foregroundViewController = toViewController;
    UIViewController *backgroundViewController = fromViewController;

    if (type == Dismissing) {
        foregroundViewController = fromViewController;
        backgroundViewController = toViewController;
    }

    //get the target view
    UIViewController *targetViewController = backgroundViewController;
    if ([targetViewController isKindOfClass:[UINavigationController class]]) {
        UINavigationController *navController = (UINavigationController *)targetViewController;
        targetViewController = navController.topViewController;
    }

    UIView *targetViewMaybe = ([targetViewController conformsToProtocol:@protocol(ExpandingTransitionPresentingViewController)])?[targetViewController expandingTransitionTargetViewForTransition:self]:assert(targetViewMaybe != nil);

现在问题就出现了，我必须在targetViewController上调用协议方法，这是一个UIViewController，但编译器提示我输入错误：

UIViewController没有可见的接口来声明选择器'expandingTransitionTargetViewForTransition'

Answer 1

问题的原因是targetViewController属于UIViewController类型。但是您尝试从某个协议调用方法。解决方案是在检查它实际符合协议后将指针强制转换为适当的类型。

帮自己一个忙，避免冗长，难以阅读，不可能像上一个那样调试行：

UIView *targetViewMaybe = ([targetViewController conformsToProtocol:@protocol(ExpandingTransitionPresentingViewController)])?[targetViewController expandingTransitionTargetViewForTransition:self]:assert(targetViewMaybe != nil);

将其拆分成可管理的东西。以下内容修复了该问题，使代码更易于阅读和调试：

if ([targetViewController conformsToProtocol:@protocol(ExpandingTransitionPresentingViewController)]) {
    id<ExpandingTransitionPresentingViewController> controller = (id<ExpandingTransitionPresentingViewController>)targetViewController;
    UIView *targetView = [controller expandingTransitionTargetViewForTransition:self];
    // Do something with targetView
} else {
    // Handle as needed
}

扩大细胞转变

1 个答案: