释放在WPF Application.OnStartUp()中创建的命名互斥体:哪个线程拥有它? [英] Releasing a named mutex created in WPF Application.OnStartUp(): Which thread owns it?

问题描述

我在WPF应用程序的OnStartup方法内创建一个互斥锁.互斥锁不在程序中的其他任何地方使用，其唯一目的是防止某些程序同时运行.当应用程序关闭时，如何释放此互斥锁?

I create a mutex within the OnStartup Method of a WPF app. The mutex is not used anywhere else in the program, its only purpose is to prevent certain programs from running concurrently. How can I release this mutex when the application closes?

根据 documentation ，mutex.ReleaseMutex()必须从创建互斥锁的同一线程中调用.但是，这带来了一个问题，因为我不控制调用OnStartup()的线程.

According to the documentation, mutex.ReleaseMutex() must be called from the same thread that created the mutex. However this presents a problem, since I do not control the thread that calls OnStartup().

假设我的OnStartup方法看起来像这样:

Suppose my OnStartup method looks like this:

public partial class App : Application
{
    private Mutex mutex;
    private bool hasHandle = false;

    protected override void OnStartup(StartupEventArgs e)
    {
        bool createdNew;
        mutex = new Mutex(false, @"Global\XYZ", out createdNew);
        try
        {
             hasHandle = mutex.WaitOne(5000, false);
             if (!hasHandle)
                 {/*do stuff*/};
        }
        catch (AbandonedMutexException)
        {
             hasHandle = true;
             // do stuff
        }
        base.OnStartup(e);
    }

    private void releaseMutex()
    {
        if (mutex!=null)
        {
             if (hasHandle) mutex.ReleaseMutex();
             mutex.Dispose();
        }
    }
}

保存以致电releaseMutex() ...

• 在OnExit()方法中?
  protected override void OnExit(){releaseMutex();}
• 是否在ProcessExit事件处理程序中?
  AppDomain.CurrentDomain.ProcessExit += (sender,e)=> releaseMutex();
• 在终结器中?
  ~App(){releaseMutex();}
• 在未处理的异常事件处理程序中?
  AppDomain.CurrentDomain.UnhandledException += (sender,e)=> releaseMutex();

• in the OnExit() method?
  protected override void OnExit(){releaseMutex();}
• in the ProcessExit event handler?
  AppDomain.CurrentDomain.ProcessExit += (sender,e)=> releaseMutex();
• in a finalizer?
  ~App(){releaseMutex();}
• in the unhandled exception event handler?
  AppDomain.CurrentDomain.UnhandledException += (sender,e)=> releaseMutex();

似乎OnExit方法有最好的机会进入相同的线程，但这似乎是一个粗略的假设.有没有办法忽略相同线程的需求?还是应该与互斥锁一起创建并存储一个单独的线程?

It seems like the OnExit method has the best chance to be in the same thread, but even that seems a sketchy assumption. Is there a way to ignore the same-thread requirement? Or should I create and store a separate thread in conjunction with my mutex?

推荐答案

我个人根本不会费心释放它，尤其是因为您处理AbandonedMutexException.

I personally wouldn't bother releasing it at all, especially since you handle AbandonedMutexException.

如果互斥锁不用于同步同一进程的线程，则无需显式释放它.当进程终止时，操作系统会自动关闭该进程创建的所有句柄，例如文件，套接字，互斥量，信号量和事件句柄.

If a mutex is not used to synchronize threads of the same process there is no need to explicitly release it. When a process terminates OS automatically closes all handles created by the process, such as files, sockets, mutexes, semaphores and event handles .

如果您仍然希望释放它，请考虑使用Application.OnExit()，因为它是从主线程中调用的，就像Startup()一样.

If you still prefer to release it consider using Application.OnExit() since it is called from the main thread, just like the Startup().

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