Java多线程同步——生产者消费者问题

这是马士兵老师的Java视频教程里的一个生产者消费者问题的模型

public class ProduceConsumer{
    public static void main(String[] args){
        SyncStack ss = new SyncStack();
        Producer pro = new Producer(ss);
        Consumer con = new Consumer(ss);
        new Thread(pro).start();
        new Thread(con).start();

    }
}

class Product{
    int id;
    public Product(int id){
        this.id = id;
    }
    public String toString(){
        return "Product:" + id;
    }
}

class SyncStack{
    int index  = 0;
    Product[] arrPro = new Product[6];

    public synchronized void push(Product p){
        while (index == arrPro.length){
            try {
                this.wait();
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

        this.notify();
        arrPro[index] = p;
        index++;
    }

    public synchronized Product pop(){
        while (index == 0){
            try {
                this.wait();
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

        this.notify();
        index--;
        return arrPro[index];
    }

}

class Producer implements Runnable{
    SyncStack ss = null;
    public Producer(SyncStack ss){ //持有SyncStack的一个引用
        this.ss = ss;
    }
    @Override
    public void run() {
        for(int i=0; i<20; i++){
            Product p = new Product(i);
            ss.push(p);
System.out.println("生产了：" + p);
            try {
                Thread.sleep(100);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }

    }
}

class Consumer implements Runnable{

    SyncStack ss = null;
    public Consumer(SyncStack ss){ //持有SyncStack的一个引用
        this.ss = ss;
    }
    @Override
    public void run() {
        for(int i=0; i<20; i++){
            Product p = ss.pop();
System.out.println("消费了：" + p);
            try {
                Thread.sleep(1000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }

    }
}