首页 > 编程 > C# > 正文

C#线程队列用法实例分析

2020-01-24 01:29:19
字体:
来源:转载
供稿:网友

本文实例讲述了C#线程队列用法。分享给大家供大家参考。具体如下:

using System;using System.Collections.Generic;using System.Text;using System.Threading;namespace ThreadPro{ class Program {  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);  static void Main(string[] args)  {   Console.WriteLine("Mutex Sample ");   //创建一个Mutex对象,并且命名为MyMutex   gM1 = new Mutex(true, "MyMutex");   //创建一个未命名的Mutex 对象.   gM2 = new Mutex(true);   Console.WriteLine(" - Main Owns gM1 and gM2");   AutoResetEvent[] evs = new AutoResetEvent[4];   evs[0] = Event1; //为后面的线程t1,t2,t3,t4定义AutoResetEvent对象   evs[1] = Event2;   Program tm = new Program();   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();// 使用Mutex.WaitAll()方法等待一个Mutex数组中的对象全部被释放   t2.Start();// 使用Mutex.WaitOne()方法等待gM1的释放   t3.Start();// 使用Mutex.WaitAny()方法等待一个Mutex数组中任意一个对象被释放   t4.Start();// 使用Mutex.WaitOne()方法等待gM2的释放   Thread.Sleep(2000);   Console.WriteLine(" - Main releases gM1");   gM1.ReleaseMutex(); //线程t2,t3结束条件满   Thread.Sleep(1000);   Console.WriteLine(" - Main releases gM2");   gM2.ReleaseMutex(); //线程t1,t4结束条件满足   //等待所有四个线程结束   WaitHandle.WaitAll(evs);   Console.WriteLine(" Mutex Sample");   Console.ReadLine();  }  public void t1Start()  {   Console.WriteLine("方法一运行, Mutex.WaitAll(Mutex[])");   Mutex[] gMs = new Mutex[2];   gMs[0] = gM1;//创建一个Mutex数组作为Mutex.WaitAll()方法的参数   gMs[1] = gM2;   Mutex.WaitAll(gMs);//等待gM1和gM2都被释放   gM1.ReleaseMutex(); //修正上一次出现的错误   gM2.ReleaseMutex(); //修正上一次出现的错误   Thread.Sleep(2000);   Console.WriteLine("方法一完毕,WaitAll(Mutex[]) satisfied");   Event1.Set(); //线程结束,将Event1设置为有信号状态  }  public void t2Start()  {   Console.WriteLine("方法二运行, gM1.WaitOne( )");   gM1.WaitOne();//等待gM1的释放   gM1.ReleaseMutex(); //修正上一次出现的错误   Console.WriteLine("方法二完毕, gM1.WaitOne( ) satisfied");   Event2.Set();//线程结束,将Event2设置为有信号状态  }  public void t3Start()  {   Console.WriteLine("t3Start started, Mutex.WaitAny(Mutex[])");   Mutex[] gMs = new Mutex[2];   gMs[0] = gM1;//创建一个Mutex数组作为Mutex.WaitAny()方法的参数   gMs[1] = gM2;   Mutex.WaitAny(gMs);//等待数组中任意一个Mutex对象被释放   gM1.ReleaseMutex(); //修正上一次出现的错误   Console.WriteLine("t3Start finished, Mutex.WaitAny(Mutex[])");   Event3.Set();//线程结束,将Event3设置为有信号状态  }  public void t4Start()  {   Console.WriteLine("t4Start started, gM2.WaitOne( )");   gM2.WaitOne();//等待gM2被释放   gM2.ReleaseMutex(); //修正上一次出现的错误   Console.WriteLine("t4Start finished, gM2.WaitOne( )");   Event4.Set();//线程结束,将Event4设置为有信号状态  } }}

希望本文所述对大家的C#程序设计有所帮助。

发表评论 共有条评论
用户名: 密码:
验证码: 匿名发表