java多线程经典问题-------生产者、消费者问题

public class ProducerAndConsumer {

	/**
	 *
title:main
	 *
function:
	 *@author dcl
	 *@param args
	 */
	
	static List
 
   hotDogs = new ArrayList
  
   (); //热狗集合 public static void main(String[] args) { for(int i = 1;i<4;i++){ new Thread(new Producer(i)).start(); } for(int j = 1;j<5;j++){ new Thread(new Consumer(j)).start(); } try { Thread.sleep(5000); System.exit(2); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } static class Producer implements Runnable{ private int pid = -1; int i = 1; //标识热狗编号 public Producer(int pid) { this.pid = pid; } @Override public void run() { /** * 生产者生产热狗，只要有热狗生产出来就可以唤醒所有的等待线程， * 热狗数量超过了展示台5个，就等待消费 */ while(true){ try { Thread.sleep(200); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); }//模拟生产消耗时间 synchronized(hotDogs){ if(hotDogs.size() < 5){ //添加热狗--生产 hotDogs.add(pid*100+i); System.out.println(pid+"号热狗师傅生产"+pid*100+i+"号热狗"); i++; //有了热狗就可以叫消费者来吃 hotDogs.notifyAll(); }else{ System.out.println("热狗拷满了，快来吃啊！"); try { hotDogs.wait(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); }//等待 } } } } } static class Consumer implements Runnable{ private int cid = -1; public Consumer(int cid) { this.cid = cid; } @Override public void run() { /** * 消费者者生产热狗，只要消费了热狗就可一唤醒所有等待进程， * 热狗数量超过了展示台5个，就等待生产 */ while(true){ try { Thread.sleep(200); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); }//模拟生产消耗时间 synchronized(hotDogs){ if(hotDogs.size() > 0){ //添加热狗--生产 System.out.println("消费者" + this.cid + "正在消费一个热狗，其编号为：" + hotDogs.remove(0)); //有了热狗就可以叫消费者来吃 hotDogs.notifyAll(); }else{ System.out.println("热狗mei了，快点做啊！"); try { hotDogs.wait(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); }//等待 } } } } } }