弹出窗口时应用程序崩溃

Question

我试图在按下ios 8中的按钮时出现弹出窗口，但是按下按钮时应用程序崩溃了。我已正确设置所有故事板并使用断点我确定这条线导致问题，即使要在popover中显示的视图控制器是FirstViewController类：

UIViewController *vc = destNav.viewControllers.firstObject;

这是崩溃：

***由于未捕获的异常终止应用程序＆＃39; NSInvalidArgumentException＆＃39;，原因：＆＃39; - [SecondViewController viewControllers]：无法识别的选择器发送到实例0x1446ea3a0＆＃39;

---

这是我这部分代码的其余部分：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
// Assuming you've hooked this all up in a Storyboard with a popover presentation style
if ([segue.identifier isEqualToString:@"popover"]) {
    UINavigationController *destNav = segue.destinationViewController;
    FirstViewController *vc = destNav.viewControllers.firstObject;

    // This is the important part
    UIPopoverPresentationController *popPC = destNav.popoverPresentationController;
    popPC.delegate = self;
}
}

- (UIModalPresentationStyle)adaptivePresentationStyleForPresentationController:(UIPresentationController *)controller {
return UIModalPresentationNone;
}

Answer 1

我解决了！对不起，我发布这个很晚了

您只需要排除

行

 FirstViewController *vc = destNav.viewControllers.firstObject;

即使所有教程都显示此行

Answer 2

尝试像这样投射视图控制器：

FirstViewController *vc = (FirstViewController *)destNav.viewControllers.firstObject;