一、問題描述
生產者消費者問題是一個典型的線程同步問題。生產者生產商品放到容器中，容器有一定的容量（只能順序放，先放后拿），消費者消費商品，當容器滿了后，生產者等待，當容器為空時，消費者等待。當生產者將商品放入容器后，通知消費者；當消費者拿走商品后，通知生產者。
二、解決方案
對容器資源加鎖，當取得鎖后，才能對互斥資源進行操作。
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{//容器采用棧，先進后出
private int index = 0;
Goods[] goods = new Goods[6];
public synchronized void push(Goods good){
while(index==goods.length){//當容器滿了，生產者等待
try {
wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
goods[index]=good;
index++;
notifyAll();//當生產者放入商品后通知消費者
}
public synchronized Goods pop(){
while(index==0){//當容器內沒有商品是等待
try {
wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
index--;
notifyAll();//當消費者消費了商品后通知生產者
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("生產了："+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("消費了："+good);
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
