C#线程队列用法实例分析
发布时间:2020-12-15 06:00:54 所属栏目:百科 来源:网络整理
导读:本篇章节讲解C#线程队列用法。供大家参考研究。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
本篇章节讲解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。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |