今天研究了一下Java多线程，顺便写了一下多线程中的经典问题—–生产者消费者经典问题，推荐一个线程链接Java基础知识回顾–线程 这个里面其实写了生产者与消费者问题，估计在后面大家看起来比较费劲，所以我提取出来再讲解一遍。
package yanning; public class ProducerConsumer { public static void main(String[] args) { Tong tong = new Tong(); Producer p = new Producer(tong); Consumer c = new Consumer(tong); new Thread(p).start(); new Thread(c).start(); } } class Baozi { int id; Baozi(int id) { this.id = id; } public String toString() { return "Baozi:" + id; } } class Tong { int index; Baozi[] BZ = new Baozi[10]; public synchronized void push(Baozi baozi) { if(index == BZ.length) { try { this.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } this.notify(); BZ[index] = baozi; index ++; } public synchronized Baozi pop() { if(index == 0) { try { this.wait(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } this.notify(); index --; return BZ[index]; } } class Producer implements Runnable{ Tong tong = null; Producer(Tong tong) { this.tong = tong; } public void run() { for( int i = 0; i < 10; i ++ ) { Baozi baozi = new Baozi( i ); tong.push(baozi); System.out.println("生产了 :" + baozi); try { Thread.sleep(1000); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } } } class Consumer implements Runnable{ Tong tong = null; Consumer(Tong tong) { this.tong = tong; } public void run() { for( int i = 0; i < 10; i ++ ) { Baozi baozi = tong.pop(); System.out.println("消费了:" + baozi); try { Thread.sleep(1000); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } } }