Question

当我尝试使用UITapGestureRecognizer时，我的iOS模拟器崩溃了。

这是我的手势：

UITapGestureRecognizer tap = new UITapGestureRecognizer (new NSAction(delegate {
  if(SettingsTapped != null){
     SettingsTapped(this, EventArgs.Empty);
  }
}));

我将此Gesture添加到UIView，然后将此视图添加到UITableViewCell。触摸View后应用程序崩溃（每次），显示没有异常。

她是模拟器日志文件的输出：

Nov  4 10:49:47 administorsmini xXxXx[11073]: assertion failed: 13E28 12B411: libsystem_sim_trace.dylib + 19982 [BEE53863-0DEC-33B1-BFFB-8F7AE595CC73]: 0x4
Nov  4 10:49:49 administorsmini xXxXx[11073]: Stacktrace:
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at <unknown> <0xffffffff>
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at (wrapper managed-to-native) MonoTouch.UIKit.UIApplication.UIApplicationMain (int,string[],intptr,intptr) <IL 0x000a6, 0xffffffff>
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at MonoTouch.UIKit.UIApplication.Main (string[],intptr,intptr) [0x00005] in /Developer/MonoTouch/Source/monotouch/src/UIKit/UIApplication.cs:62
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at MonoTouch.UIKit.UIApplication.Main (string[],string,string) [0x00038] in /Developer/MonoTouch/Source/monotouch/src/UIKit/UIApplication.cs:46
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at xXxXx.Application.Main (string[]) [0x00008] in /Users/Norman/Desktop/xXxXx/xXxXx/Main.cs:17
Nov  4 10:49:49 administorsmini xXxXx[11073]:   at (wrapper runtime-invoke) <Module>.runtime_invoke_void_object (object,intptr,intptr,intptr) <IL 0x00050, 0xffffffff>
Nov  4 10:49:49 administorsmini xXxXx[11073]: 
Native stacktrace:
Nov  4 10:49:49 administorsmini xXxXx[11073]: 
=================================================================
Got a SIGSEGV while executing native code. This usually indicates
a fatal error in the mono runtime or one of the native libraries 
used by your application.
=================================================================
Nov  4 10:49:49 administorsmini com.apple.CoreSimulator.SimDevice.27B5D497-B641-4BCA-8FA0-EF9E28E07143.launchd_sim[10951] (UIKitApplication:com.your-company.xXxXx[0x7041][11073]): Service exited due to signal: Abort trap: 6
Nov  4 10:49:49 administorsmini SpringBoard[10962]: Application 'UIKitApplication:com.your-company.xXxXx[0x7041]' crashed.
Nov  4 10:49:49 administorsmini assertiond[10966]: notify_suspend_pid() failed with error 7
Nov  4 10:49:49 administorsmini assertiond[10966]: assertion failed: 13E28 12B411: assertiond + 11523 [3F572A0B-7E12-378D-AFEE-EA491BAF2C36]: 0x1

我现在能做什么？我不想在设备上开发......

修改

我最终通过覆盖UIView并使用其TouchesBegan()方法。有许多方法可以点击某些东西，但为什么我不能在上面使用这个呢？

Answer 1

这是我在我的应用程序中使用的一些代码：

private void addTapGesture()
{
   imgLogo.UserInteractionEnabled = true;
   var tapGesture = new UITapGestureRecognizer(this, new Selector("ResendTrigger:"));
   tapGesture.NumberOfTapsRequired = 5;
   imgLogo.AddGestureRecognizer(tapGesture);           
}

[Export("ResendTrigger:")]
public void ResendTrigger(UIGestureRecognizer sender)
{
    System.Diagnostics.Debug.WriteLine("Triggered");}
}

构造函数的第一个参数（this）指向包含具有指定Export属性的方法的定义的对象，因此如果您将在视图中定义选择器和方法，如果您要调用的方法存在，则NSObject目标将引用您的视图单元格中的示例您的单元格引用将成为目标。

编辑

根据评论我假设您可能正在使用UITableViewSource。我认为你的情况与我的相似，所以尝试这种方法：

在你的UITableSource中声明事件类似于public event EventHandler<DataBaseModels.SavedActions> OnActionSelected;的事件然后在getCell中为你的视图分配点击手势在内部方法中，您将引用发件人objec，将其强制转换为UIView并保留标记以识别相应的记录

UITapGestureRecognizer在iOS模拟器上崩溃

1 个答案: