swift prepareForSegue 不工作 / exc_breakpoint (code=exc_i386_bpt subcode=0x0)答案

【问题标题】：swift prepareForSegue not working / exc_breakpoint (code=exc_i386_bpt subcode=0x0)swift prepareForSegue 不工作 / exc_breakpoint (code=exc_i386_bpt subcode=0x0)
【发布时间】：2014-12-23 03:22:46
【问题描述】：

我有这个根本没有被调用的函数。我没有准备好继续打印...

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？

【问题讨论】：

标签： ios swift

【解决方案1】：

您的 ChatRoomViewController 是否位于导航控制器中？如果是，那么 segue.destinationViewController 不会指向 ChatRoom 控制器？只是一个猜测。在这种情况下，您需要类似：

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。从情节提要中显式实例化的下一行绕过了对它的需求。

【讨论】：