为什么Microsoft的示例代码崩溃了？

我想了解一下.NET中的multithreading，并从MSDN中获取此示例。 它编译很好但在运行时崩溃。 我希望微软会告诉我创建多个线程的正确方法。 我无法理解为什么它崩溃了。 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

您正在关注VS2003 / .NET 1.1时代的教程。

AbandonedMutexException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Wait函数）获得的相应互斥锁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锁的情况下退出，则下一个等待的线程将获得该互斥锁，就好像没有发生任何不良事件一样。 但是，由于这经常导致不正确的行为（例如，在更新受互斥锁保护的状态时退出线程），因此对BCL进行了更改。

说实话，编写基于Mutex的代码很少（如果有的话），您想要为大多数真正的单词线程场景做些什么。 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实际情况，并且您认为互斥锁可能是解决方案，我建议您提出一个包含详细信息的单独问题。

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在退出之前重写4个线程函数以释放相应的互斥锁。 唯一真正棘手的是t3（计算得到的互斥量）：

 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(); }