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();
			}
		
		}
		
	}	
}