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第九章 Java多线程机制 05_线程同步_7_生产者消费者问题

2012-09-03 23:25 585 查看
鱼欲遇雨:每日都学习一点,持之以恒,天道酬勤!不能用电脑时,提前补上!(2012.9.3)

模拟消费者和生产者之间的关系

示例代码:(线程的核心)

// ProducerConsumer.java

//import java.util.Math;

public class ProducerConsumer {
public static void main(String args[]) {
SyncStack ss = new SyncStack();
Producer p = new Producer(ss);
Consumer c = new Consumer(ss);
new Thread(p).start();
new Thread(c).start();
}
}

class WoTou {
int id;
WoTou(int id) {
this.id = id;
}

public String toString() {
return ("WoTou: " + id);
}
}

class SyncStack {
int index;
WoTou[] arrWT = new WoTou[6];

public synchronized void  push(WoTou wt) {
while(index == 6) {
try
{
this.wait();    //执行当前对象的线程等待;必须有锁定才能wait();wait()不在锁定
}
catch (InterruptedException e)
{
e.printStackTrace();
}
}
this.notify();    //叫醒当前等待的线程
arrWT[index] = wt;
index ++;
}

public synchronized WoTou pop() {
while(index == 0) {
try
{
this.wait();
}
catch (InterruptedException e)
{
e.printStackTrace();
}
}
this.notify();
index --;
return arrWT[index];
}
}

class Producer implements Runnable {
SyncStack ss = null;

Producer(SyncStack ss) {
this.ss = ss;
}

public void run() {
for(int i = 0; i < 20; i ++) {
WoTou wt = new WoTou(i);
ss.push(wt);
System.out.println("生产了: " + wt);
try
{
Thread.sleep((int)(Math.random()  * 200));
}
catch (InterruptedException e)
{
e.printStackTrace();
}
}
}
}

class Consumer implements Runnable {
SyncStack ss = null;

Consumer(SyncStack ss) {
this.ss = ss;
}

public void run() {
for(int i = 0; i < 20; i ++) {
WoTou wt = ss.pop();
System.out.println("消费了: " + wt);
try
{
Thread.sleep((int)(Math.random() * 1000));
}
catch (InterruptedException e)
{
e.printStackTrace();
}
}
}
}
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