多线程之线程通信(管程法,信号灯法)

1.生产者消费者模式

代码:
1.管程法解决生产者消费者问题

package com.qiu.syn;
//测试生产者消费者-->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 synContainer = new SynContainer();
        new Productor(synContainer).start();
        new Consumer(synContainer).start();
    }
}
//生产者
class Productor extends Thread{
     
    SynContainer container;
    public Productor(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;
​    }
​    @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;//产品编号

​    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];
    //生产者放入产品
    //容器计数器
    int count = 0;
    public synchronized void push(Chicken chicken){
     
        //如果容器满了
        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(){
     
        //判断是否能够消费
        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];
        //通话生产者生产
        this.notifyAll();
        return chicken;
    }
}

1.信号灯法解决(标志位解决)
代码:

package com.qiu.syn;

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("天天向上");
​            }else {
     
​                this.tv.play("王牌对王牌,有胆你就来");
​            }
​        }
​    }
}
//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;
​    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("观众观看了:"+voice);

​        //通知演员表演
​        this.notifyAll();//通知唤醒
​        this.flag = !this.flag;
​    }
}