为什么 Microsoft 的这个示例代码会崩溃？

Question

我想了解一些有关 .NET 中多线程的知识，并从 MSDN。它编译得很好，但在运行时崩溃。我希望微软能告诉我创建多线程的正确方法。我不明白为什么它会崩溃。有人可以帮忙吗？

// Mutex.cs
// Mutex object example
using System;
using System.Threading;

public class MutexSample
{
   static Mutex gM1;
   static Mutex gM2;
   const int ITERS = 100;
   static AutoResetEvent Event1 = new AutoResetEvent(false);
   static AutoResetEvent Event2 = new AutoResetEvent(false);
   static AutoResetEvent Event3 = new AutoResetEvent(false);
   static AutoResetEvent Event4 = new AutoResetEvent(false);

   public static void Main(String[] args)
   {
      Console.WriteLine("Mutex Sample ...");

      // Create Mutex initialOwned, with name of "MyMutex".
      gM1 = new Mutex(true,"MyMutex");
      // Create Mutex initialOwned, with no name.
      gM2 = new Mutex(true);

      Console.WriteLine(" - Main Owns gM1 and gM2");

      AutoResetEvent[] evs = new AutoResetEvent[4];
      evs[0] = Event1;    // Event for t1
      evs[1] = Event2;    // Event for t2
      evs[2] = Event3;    // Event for t3
      evs[3] = Event4;    // Event for t4

      MutexSample tm = new MutexSample( );
      Thread t1 = new Thread(new ThreadStart(tm.t1Start));
      Thread t2 = new Thread(new ThreadStart(tm.t2Start));
      Thread t3 = new Thread(new ThreadStart(tm.t3Start));
      Thread t4 = new Thread(new ThreadStart(tm.t4Start));

      t1.Start( );   // Does Mutex.WaitAll(Mutex[] of gM1 and gM2)
      t2.Start( );   // Does Mutex.WaitOne(Mutex gM1)
      t3.Start( );   // Does Mutex.WaitAny(Mutex[] of gM1 and gM2)
      t4.Start( );   // Does Mutex.WaitOne(Mutex gM2)

      Thread.Sleep(2000);
      Console.WriteLine(" - Main releases gM1");
      gM1.ReleaseMutex( );  // t2 and t3 will end and signal

      Thread.Sleep(1000);
      Console.WriteLine(" - Main releases gM2");
      gM2.ReleaseMutex( );  // t1 and t4 will end and signal

      // Waiting until all four threads signal that they are done.
      WaitHandle.WaitAll(evs); 
      Console.WriteLine("... Mutex Sample");
   }

   public void t1Start( )
   {
      Console.WriteLine("t1Start started,  Mutex.WaitAll(Mutex[])");
      Mutex[] gMs = new Mutex[2] { gM1, gM2};
      Mutex.WaitAll(gMs);  // Waits until both gM1 and gM2 are released
      Thread.Sleep(2000);
      Console.WriteLine("t1Start finished, Mutex.WaitAll(Mutex[]) satisfied");
      Event1.Set( );      // AutoResetEvent.Set() flagging method is done
   }

   public void t2Start( )
   {
      Console.WriteLine("t2Start started,  gM1.WaitOne( )");
      gM1.WaitOne( );    // Waits until Mutex gM1 is released
      Console.WriteLine("t2Start finished, gM1.WaitOne( ) satisfied");
      Event2.Set( );     // AutoResetEvent.Set() flagging method is done
   }

   public void t3Start( )
   {
      Console.WriteLine("t3Start started,  Mutex.WaitAny(Mutex[])");
      Mutex[] gMs = new Mutex[2] { gM1, gM2};
      Mutex.WaitAny(gMs);  // Waits until either Mutex is released
      Console.WriteLine("t3Start finished, Mutex.WaitAny(Mutex[])");
      Event3.Set( );       // AutoResetEvent.Set() flagging method is done
   }

   public void t4Start( )
   {
      Console.WriteLine("t4Start started,  gM2.WaitOne( )");
      gM2.WaitOne( );   // Waits until Mutex gM2 is released
      Console.WriteLine("t4Start finished, gM2.WaitOne( )");
      Event4.Set( );    // AutoResetEvent.Set() flagging method is done
   }
}

Answer 1

您正在学习 VS2003/.NET 1.1 时代的教程。

AbandonedMutexException 是在 .NET 2 中引入的，因此为什么如果在 .NET 2 或更高版本上使用，示例代码现在会失败。

每个线程（通过等待函数）获得的适当互斥体应在线程退出之前释放。在 1.1 代码中，如果一个线程在拥有互斥体的同时退出，则下一个等待线程将获得该互斥体，就好像没有发生任何意外一样。然而，由于这经常导致不正确的行为（例如，在更新互斥锁保护的状态时线程退出），因此对 BCL 进行了更改。

老实说，对于大多数实际的字线程场景来说，编写基于互斥锁的代码很少（如果有的话）。如果这只是一个学习练习，我建议忽略它们。如果您正在尝试解决现实场景，并且您认为互斥体可能是解决方案，那么我建议您提出一个包含详细信息的单独问题。

---

重写了 4 个线程函数，以便在退出之前释放适当的互斥锁。唯一真正棘手的是 t3（计算出获得了哪个互斥体）：

public void t1Start()
{
    Console.WriteLine("t1Start started,  Mutex.WaitAll(Mutex[])");
    Mutex[] gMs = new Mutex[2] { gM1, gM2 };
    Mutex.WaitAll(gMs);  // Waits until both gM1 and gM2 are released
    Thread.Sleep(2000);
    Console.WriteLine("t1Start finished, Mutex.WaitAll(Mutex[]) satisfied");
    Event1.Set();      // AutoResetEvent.Set() flagging method is done
    gM1.ReleaseMutex();
    gM2.ReleaseMutex();
}

public void t2Start()
{
    Console.WriteLine("t2Start started,  gM1.WaitOne( )");
    gM1.WaitOne();    // Waits until Mutex gM1 is released
    Console.WriteLine("t2Start finished, gM1.WaitOne( ) satisfied");
    Event2.Set();     // AutoResetEvent.Set() flagging method is done
    gM1.ReleaseMutex();
}

public void t3Start()
{
    Console.WriteLine("t3Start started,  Mutex.WaitAny(Mutex[])");
    Mutex[] gMs = new Mutex[2] { gM1, gM2 };
    int mxObtained = Mutex.WaitAny(gMs);  // Waits until either Mutex is released
    Console.WriteLine("t3Start finished, Mutex.WaitAny(Mutex[])");
    Event3.Set();       // AutoResetEvent.Set() flagging method is done
    gMs[mxObtained].ReleaseMutex();
}

public void t4Start()
{
    Console.WriteLine("t4Start started,  gM2.WaitOne( )");
    gM2.WaitOne();   // Waits until Mutex gM2 is released
    Console.WriteLine("t4Start finished, gM2.WaitOne( )");
    Event4.Set();    // AutoResetEvent.Set() flagging method is done
    gM2.ReleaseMutex();
}