为什么这个来自Microsoft的示例代码会崩溃

本文关键字：代码 崩溃 Microsoft 为什么 | 更新日期: 2023-09-27 18:00:29

我想了解一些.NET中的多线程技术，并从MSDN中获得了这个示例。它编译良好，但在运行时崩溃。我希望微软能告诉我创建多个线程的正确方法。我不知道它为什么会崩溃。有人能帮忙吗？

// 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
   }
}

为什么这个来自Microsoft的示例代码会崩溃

您正在学习VS2003/.NET 1.1时代的教程。

放弃的MutexException是在.NET 2中引入的，因此，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）获得的适当互斥应该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锁的情况下退出，那么下一个等待的线程就会获得互斥锁，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，线程在更新受互斥体保护的状态时退出），因此对BCL进行了更改。

老实说，在大多数真实的字线程场景中，编写基于Mutex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。如果你试图解决现实世界中的场景，并且你认为互斥可能是解决方案，我建议你单独问一个包含细节的问题。

重新编写4个线程函数，以便在退出之前释放相应的互斥对象。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();
}
相关文章：