ios - swift prepareForSegue 不工作/exc_breakpoint(代码=exc_i386_bpt 子代码=0x0)

Question

我有一个根本没有被调用的函数。我什至没有打印 segue 的准备...

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
        println("PREPARING FOR SEGUE");
        if (segue.identifier == "ToChatRoom") {
            var chatView:ChatRoomViewController = segue.destinationViewController as ChatRoomViewController;
            var index = coralReefTableView.indexPathForSelectedRow()!.row;
            var id = roomIDArray.objectAtIndex(index);
            println("ID IS : \(id)");
            chatView.selectedRoomID = id as Int;
        }
    }

我正在用这些代码行调用 segue...

    func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {
        var cell = tableView.cellForRowAtIndexPath(indexPath);
        //performSegueWithIdentifier("ToChatRoom", sender: self);
        let nextController:AnyObject! = self.storyboard?.instantiateViewControllerWithIdentifier("chatRoom");
        self.showViewController(nextController as UIViewController, sender: nextController);

当我取消注释 performSegueWithIdentifier 时，我收到此错误:exc_breakpoint (code=exc_i386_bpt subcode=0x0)。我想知道这是什么原因造成的？

Answer 1

您的 ChatRoomViewController 是否在导航 Controller 中？如果是，那么 segue.destinationViewController 不会指向 ChatRoom Controller ？只是一个猜测。在这种情况下，您需要类似的东西:

if segue.identifier == "ToChatRoom" {
   let navigationController = segue.destinationViewController as UINavigationController
   let chatView = navigationController.viewControllers[0] as ChatRoomViewController
   var index = ...

当 performSegueWithIdentifier 行被注释掉时，prepareForSegue 永远不会被调用。从 Storyboard 中显式实例化的下一行绕过了对它的需要。