另一个解决方案依赖于responder链，它可能工作，也可能不工作，这取决于第一个responder是什么:

获取第一个响应器。 获取与第一个responder相关联的UIViewController。

示例伪代码:

+ (UIViewController *)currentViewController {
    UIView *firstResponder = [self firstResponder]; // from the first link above, but not guaranteed to return a UIView, so this should be handled more appropriately.
    UIViewController *viewController = [firstResponder viewController]; // from the second link above
    return viewController;
}

iOS 4在UIWindow上引入了rootViewController属性:

[UIApplication sharedApplication].keyWindow.rootViewController;

你需要在创建视图控制器后自己设置。

你应该使用:

[UIApplication sharedApplication].window.rootViewController;

当[UIApplication sharedApplication]上有一个uiactionsheet时。keyWindow，这是不正确的使用keyWindow在这个答案中提到。

Swift替代解决方案:

static func topMostController() -> UIViewController {
    var topController = UIApplication.sharedApplication().keyWindow?.rootViewController
    while (topController?.presentedViewController != nil) {
        topController = topController?.presentedViewController
    }

    return topController!
}

为了完成Eric的回答(谁省略了弹出窗口，导航控制器，选项卡控制器，视图控制器添加为子视图到一些其他的视图控制器遍历)，这是我的版本返回当前可见的视图控制器:

=====================================================================

- (UIViewController*)topViewController {
    return [self topViewControllerWithRootViewController:[UIApplication sharedApplication].keyWindow.rootViewController];
}

- (UIViewController*)topViewControllerWithRootViewController:(UIViewController*)viewController {
    if ([viewController isKindOfClass:[UITabBarController class]]) {
        UITabBarController* tabBarController = (UITabBarController*)viewController;
        return [self topViewControllerWithRootViewController:tabBarController.selectedViewController];
    } else if ([viewController isKindOfClass:[UINavigationController class]]) {
        UINavigationController* navContObj = (UINavigationController*)viewController;
        return [self topViewControllerWithRootViewController:navContObj.visibleViewController];
    } else if (viewController.presentedViewController && !viewController.presentedViewController.isBeingDismissed) {
        UIViewController* presentedViewController = viewController.presentedViewController;
        return [self topViewControllerWithRootViewController:presentedViewController];
    }
    else {
        for (UIView *view in [viewController.view subviews])
        {
            id subViewController = [view nextResponder];
            if ( subViewController && [subViewController isKindOfClass:[UIViewController class]])
            {
                if ([(UIViewController *)subViewController presentedViewController]  && ![subViewController presentedViewController].isBeingDismissed) {
                    return [self topViewControllerWithRootViewController:[(UIViewController *)subViewController presentedViewController]];
                }
            }
        }
        return viewController;
    }
}

=====================================================================

现在你需要做的就是像下面这样调用上面的方法:

UIViewController *topMostViewControllerObj = [self topViewController];

这是对Eric的回答的改进:

UIViewController *_topMostController(UIViewController *cont) {
    UIViewController *topController = cont;

    while (topController.presentedViewController) {
        topController = topController.presentedViewController;
    }

    if ([topController isKindOfClass:[UINavigationController class]]) {
        UIViewController *visible = ((UINavigationController *)topController).visibleViewController;
        if (visible) {
            topController = visible;
        }
    }

    return (topController != cont ? topController : nil);
}

UIViewController *topMostController() {
    UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;

    UIViewController *next = nil;

    while ((next = _topMostController(topController)) != nil) {
        topController = next;
    }

    return topController;
}

UIViewController *cont是一个辅助函数。

现在你所需要做的就是调用topMostController()和最顶端的UIViewController应该被返回!