JAVA生产者消费者(线程同步)代码学习示例
一、问题描述
生产者消费者问题是一个典型的线程同步问题。生产者生产商品放到容器中,容器有一定的容量(只能顺序放,先放后拿),消费者消费商品,当容器满了后,生产者等待,当容器为空时,消费者等待。当生产者将商品放入容器后,通知消费者;当消费者拿走商品后,通知生产者。
二、解决方案
对容器资源加锁,当取得锁后,才能对互斥资源进行操作。
public class ProducerConsumerTest {
public static void main(String []args){
Container con = new Container();
Producer p = new Producer(con);
Consumer c = new Consumer(con);
new Thread(p).start();
new Thread(c).start();
}
}
class Goods{
int id;
public Goods(int id){
this.id=id;
}
public String toString(){
return "商品"+this.id;
}
}
class Container{//容器采用栈,先进后出
private int index = 0;
Goods[] goods = new Goods[6];
public synchronized void push(Goods good){
while(index==goods.length){//当容器满了,生产者等待
try {
wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
goods[index]=good;
index++;
notifyAll();//当生产者放入商品后通知消费者
}
public synchronized Goods pop(){
while(index==0){//当容器内没有商品是等待
try {
wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
index--;
notifyAll();//当消费者消费了商品后通知生产者
return goods[index];
}
}
class Producer implements Runnable{
Container con = new Container();
public Producer(Container con){
this.con=con;
}
public void run(){
for(int i=0; i<20; i++){
Goods good = new Goods(i);
con.push(good);
System.out.println("生产了:"+good);
try {
Thread.sleep(100);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
class Consumer implements Runnable{
Container con = new Container();
public Consumer(Container con){
this.con=con;
}
public void run(){
for(int i=0; i<20; i++){
Goods good=con.pop();
System.out.println("消费了:"+good);
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}