• 线程池


    引用:http://www.west263.com/www/info/22460-1.htm

    任何游戏都至少需要运行两个线程,主线程和gui线程 
    而线程池是一个管理运行线程的有用工具,下面的代码示范了一个线程池的实现方法~~ 
    ************************************************ 
    (threadpool.java) 
    import java.util.linkedlist; 

    /** 
    线程池是一组线程,限制执行任务的线程数 
    */ 
    public class threadpool extends threadgroup { 

    private boolean isalive; 
    private linkedlist taskqueue; 
    private int threadid; 
    private static int threadpoolid; 

    /** 
    创建新的线程池,numthreads是池中的线程数 
    */ 
    public threadpool(int numthreads) { 
    super("threadpool-" + (threadpoolid++)); 
    setdaemon(true); 

    isalive = true; 

    taskqueue = new linkedlist(); 
    for (int i=0; i<numthreads; i++) { 
    new pooledthread().start(); 


    /** 
    请求新任务。人物在池中下一空闲线程中运行,任务按收到的顺序执行 
    */ 
    public synchronized void runtask(runnable task) { 
    if (!isalive) { 
    throw new illegalstateexception();//线程被关则抛出illegalstateexception异常 

    if (task != null) { 
    taskqueue.add(task); 
    notify(); 





    protected synchronized runnable gettask() 
    throws interruptedexception 

    while (taskqueue.size() == 0) { 
    if (!isalive) { 
    return null; 

    wait(); 

    return (runnable)taskqueue.removefirst(); 



    /** 
    关闭线程池,所有线程停止,不再执行任务 
    */ 
    public synchronized void close() { 
    if (isalive) { 
    isalive = false; 
    taskqueue.clear(); 
    interrupt(); 




    /** 
    关闭线程池并等待所有线程完成,执行等待的任务 
    */ 
    public void join() { 
    //告诉等待线程线程池已关 
    synchronized (this) { 
    isalive = false; 
    notifyall(); 


    // 等待所有线程完成 
    thread[] threads = new thread[activecount()]; 
    int count = enumerate(threads); 
    for (int i=0; i<count; i++) { 
    try { 
    threads[i].join(); 

    catch (interruptedexception ex) { } 




    /** 
    用于进行任务的线程 
    */ 
    private class pooledthread extends thread { 


    public pooledthread() { 
    super(threadpool.this, 
    "pooledthread-" + (threadid++)); 



    public void run() { 
    while (!isinterrupted()) { 

    // 得到任务 
    runnable task = null; 
    try { 
    task = gettask(); 

    catch (interruptedexception ex) { } 

    // 若gettask()返回null或中断,则关闭此线程并返回 
    if (task == null) { 
    return; 


    // 运行任务,吸收异常 
    try { 
    task.run(); 

    catch (throwable t) { 
    uncaughtexception(this, t); 





    ********************************************* 
    要测试这个线程池,可以通过下面这个test程序! 
    ********************************************* 
    (threadpooltest.java) 
    public class threadpooltest { 

    public static void main(string[] args) { 
    if (args.length != 2) { 
    system.out.println("tests the threadpool task."); 
    system.out.println( 
    "usage: java threadpooltest numtasks numthreads"); 
    system.out.println( 
    " numtasks - integer: number of task to run."); 
    system.out.println( 
    " numthreads - integer: number of threads " + 
    "in the thread pool."); 
    return; 

    int numtasks = integer.parseint(args[0]); 
    int numthreads = integer.parseint(args[1]); 

    // 生成线程池 
    threadpool threadpool = new threadpool(numthreads); 

    // 运行任务 
    for (int i=0; i<numtasks; i++) { 
    threadpool.runtask(createtask(i)); 


    // 关闭线程池并等待所有任务完成 
    threadpool.join(); 



    /** 
    一个简单的任务(打印id) 
    */ 
    private static runnable createtask(final int taskid) { 
    return new runnable() { 
    public void run() { 
    system.out.println("task " + taskid + ": start"); 

    // 增加耗时 
    try { 
    thread.sleep(500); 

    catch (interruptedexception ex) { } 

    system.out.println("task " + taskid + ": end"); 

    }; 



    ****************************************************** 
    这样的线程池可以在许多地方应用! 

  • 相关阅读:
    许可管理工具
    浅谈MapControl控件和PageLayoutControl控件
    通过Bresenham算法实现完成矢量线性多边形向栅格数据的转化
    四叉树算法原理与实现
    OC系列foundation Kit基础-NSNumber
    OC系列foundation Kit基础-NSdictionary
    OC系列foundation Kit基础-NSMutableArray
    OC系列foundation Kit基础-NSArray
    OC系列foundation Kit基础-NSMutableString
    OC系列foundation Kit基础-NSString
  • 原文地址:https://www.cnblogs.com/sode/p/2382258.html
Copyright © 2020-2023  润新知