线程协作

生产者消费者模式

• 管程法

package com.deng.gaoji;

//测试生产者消费者模型--》利用缓冲区解决：管程法

//生产者，消费者，产品，缓冲区
public class TestPC {
    public static void main(String[] args) {
        SynContainer container=new SynContainer();

        new Producer(container).start();
        new Consumer(container).start();
    }
}

//生产者
class Producer extends Thread {
    SynContainer container;

    public Producer(SynContainer container){
        this.container=container;
    }

    //生产
    @Override
    public void run() {
        for (int i = 0; i < 100; i++) {
            container.push(new Chicken(i));
            System.out.println("生产了"+i+"只鸡");

        }
    }
}

//消费者
class Consumer extends Thread {
    SynContainer container;

    public Consumer(SynContainer container){
        this.container=container;
    }

    //消费
    @Override
    public void run() {
        for (int i = 0; i < 100; i++) {
            System.out.println("消费了-->"+container.pop().id+"只鸡");
        }
    }
}

//产品
class Chicken {
    int id;

    public Chicken(int id) {
        this.id = id;
    }
}

//缓冲区
class SynContainer {
    //需要一个容器大小
    Chicken[] chickens = new Chicken[10];
    //容器计数器
    int count = 0;

    //生产者放入产品
    public synchronized void push(Chicken chicken) {
        //如果容器满了，就需要等待消费者消费
        if (count == chickens.length) {
            //通知消费者消费，生产者等待
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

        //如果没有满，我们就需要丢入产品
        chickens[count] = chicken;
        count++;

        //可以通知消费者消费了
        this.notifyAll();
    }

    //消费者消费产品
    public synchronized Chicken pop() {
        //判断能否消费
        if (count == 0) {
            //等待生产者生产，消费者等待
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

        //如果可以消费
        count--;
        Chicken chicken = chickens[count];

        //吃完了，通知生产者生产
        this.notifyAll();
        return chicken;
    }
}

• 信号灯法

package com.deng.gaoji;

//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("广告：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 T
    //观众观看，演员等待 F
    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;
    }

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