信号灯法

//测试生产者消费者问题2：信号灯法，标志位解决

public class TestPc2 {
    public static void main(String[] args) {
        TV tv = new TV();
        new Player(tv).start();
        new Watcher(tv).start();
    }

}

//生产者-->演员
class Player extends Thread{
    TV tv;
    public Player(TV tv){
        this.tv = tv;
    }

    @Override
    public void run() {
        for (int i = 0; i < 20; i++) {
            if (i%2==0){
                this.tv.play("快乐大本营播放中");
            }else {
                this.tv.play("抖音：记录美好生活");
            }
        }
    }
}

//消费者-->观众
class Watcher extends Thread{
    TV tv;
    public Watcher(TV tv){
        this.tv = tv;
    }

    @Override
    public void run() {
        for (int i = 0; i < 20; i++) {
            tv.watch();
        }
    }
}

//产品-->节目
class TV{
    //演员表演，观众等待
    //观众观看，演员等待
    String voice;//表演的节目
    boolean flag = true;

    //表演
    public synchronized void play(String voice){

        if (!flag){
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

        System.out.println("演员表演了："+voice);
        //通知观众观看
        this.notifyAll();//通知唤醒
        this.voice = voice;
        this.flag = !this.flag;
    }

    //观看
    public synchronized void watch(){
        if (flag){
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println("观看了："+voice);
        //通知演员表演
        this.notifyAll();
        this.flag = !this.flag;
    }

}

信号灯法

原文：https://www.cnblogs.com/helloxiaolu/p/13324690.html