买票问题

最新推荐文章于 2022-06-30 19:30:13 发布

转载最新推荐文章于 2022-06-30 19:30:13 发布 · 107 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/fly-book/p/11439328.html

本文介绍了一个基于Java的多线程售票系统实现，通过synchronized关键字和对象锁确保线程安全，演示了如何控制两个窗口同时出售10张票而不出现重复售出的情况。文章提供了完整的代码示例，展示了如何在多线程环境下使用锁来避免竞态条件。


/**
 * 买票问题
 */
public class ThreadTrain1 implements Runnable {
    private int count = 10;
    private static Object o = new Object();
    @Override
    public void run() {
        while (count>0){
            try {
                Thread.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            sale();
        }
    }

    private void sale() {
        // 前提 多线程进行使用、多个线程只能拿到一把锁。
        // 保证只能让一个线程 在执行 缺点效率降低
        synchronized (o){
            if (count>0){
                System.out.println(Thread.currentThread().getName()+",出售第"+(10-count+1)+"张票");
                count--;
            }
        }
    }
    public static void main(String[] args){
        ThreadTrain1 train1 = new ThreadTrain1();
        Thread t1 = new Thread(train1,"1号窗口");
        Thread t2 = new Thread(train1,"2号窗口");
        t1.start();
        t2.start();
    }
}


/**
 * 多个线程想要同步，必须用同一把锁
 */
public class ThreadTrain1 implements Runnable {
    private int count = 10;
    private static Object o = new Object();
    public boolean flag = true;
    @Override
    public void run() {
        if (flag){
            synchronized (this){
                while (count>0){
                    try {
                        Thread.sleep(10);
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    if (count>0){
                        System.out.println(Thread.currentThread().getName()+",出售第"+(10-count+1)+"张票");
                        count--;
                    }
                }
            }
        } else {
            sale();
        }
    }

    private synchronized void sale() { //非静态同步函数用的是this锁
        while (count>0){
            if (count>0){
                try {
                    Thread.sleep(10);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName()+",出售第"+(10-count+1)+"张票");
                count--;
            }
        }
    }
    public static void main(String[] args) throws InterruptedException{
        ThreadTrain1 train1 = new ThreadTrain1();
        Thread t1 = new Thread(train1,"1号窗口");
        Thread t2 = new Thread(train1,"2号窗口");
        t1.start();
        Thread.sleep(40);
        train1.flag = false;
        t2.start();
    }
}

转载于:https://www.cnblogs.com/fly-book/p/11439328.html