麻烦大家解释一下代码Queue谢谢.

[5ihuyang]

实现要求:生产者通过Queue把生产的东西给消费者,消费者通过Queue获取生产者的信息.
代码如下:
package lizi;

class Test
{
public static void main(String[] args)
{
Queue q=new Queue();
Producer p=new Producer(q);
Consumer c=new Consumer(q);
p.start();
c.start();
}
}

class Producer extends Thread
{
Queue q;
Producer(Queue q)
{
this.q=q;
}
public void run()
{
for(int i=0;i<10;i++)
{
q.put(i);
System.out.println("Producer put "+i);
}
}
}
class Consumer extends Thread
{
Queue q;
Consumer(Queue q)
{
this.q=q;
}
public void run()
{
while(true)
{
System.out.println("Consumer get "+q.get());
}
}
}
class Queue
{
int value;
boolean bFull=false;
public synchronized void put(int i)
{
if(!bFull)
{
value=i;
bFull=true;
notify();
}
try
{
wait();
}
catch(Exception e)
{
e.printStackTrace();
}

}
public synchronized int get()
{
if(!bFull)
{
try
{
wait();
}
catch(Exception e)
{
e.printStackTrace();
}
}
bFull=false;
notify();
return value;
}
}



执行结果如下:


Producer put 0
Consumer get 0
Producer put 1
Consumer get 1
Producer put 2
Consumer get 2
Consumer get 3//出现错误现象,先get
Producer put 3
Consumer get 4//error
Producer put 4
Producer put 5
Consumer get 5
Producer put 6
Consumer get 6
Consumer get 7
Producer put 7
Consumer get 8
Producer put 8
Consumer get 9
Producer put 9

上面的代码在配置低机器上运行取得正确结果.

********************************************************
修改后的代码:

package lizi;
class Test
{
public static void main(String[] args)
{
Queue q=new Queue();
Producer p=new Producer(q);
Consumer c=new Consumer(q);
p.start();
c.start();
}
}

class Producer extends Thread
{
Queue q;
Producer(Queue q)
{
this.q=q;
}
public void run()
{
for(int i=0;i<10;i++)
{
q.put(i);


System.out.println("Producer put "+i);
}
}
}
class Consumer extends Thread
{
Queue q;
Consumer(Queue q)
{
this.q=q;
}
public void run()
{
while(true)
{
System.out.println("Consumer get "+q.get());
}
}
}
class Queue
{
int value;
boolean bFull=false;
public synchronized void put(int i)
{
if(!bFull)
{
value=i;
bFull=true;
notify();
}
try
{
wait();
}
catch(Exception e)
{
e.printStackTrace();
}

}
public synchronized int get()
{
if(!bFull)
{
try
{
wait();
}
catch(Exception e)
{
e.printStackTrace();
}
}
bFull=false;
notify();
try{ //修改的地方
Thread.sleep(1);
}
catch(Exception e)
{
e.printStackTrace();
}
return value;
}
}

运行结果
Producer put 0
Consumer get 0
Producer put 1
Consumer get 1
Producer put 2
Consumer get 2
Producer put 3
Consumer get 3
Producer put 4
Consumer get 4
Producer put 5
Consumer get 5
Producer put 6
Consumer get 6
Producer put 7
Consumer get 7
Producer put 8
Consumer get 8
Producer put 9
Consumer get 9

取得正确结果,想请教为什么上面没有修改的代码put,跟get抢cpu没有全抢?而是有的出现错误,有的没有.而在慢机器上运行正常,机器配置跟结果有多大关系?
请教高手.谢谢

[5ihuyang]

高手帮帮忙啊

[lei198203]

其实这个程序没有问题, cuonsumer get总是在producer put之后才会执行.

q.put(i);//--1
System.out.println("Producer put "+i);//--2

System.out.println("Consumer get "+q.get());//--3

在执行1的时候,如果Queue里的bFull等于true,也就是有产品可取的时候,q.get()才会返回结果,
3语句才得以执行.但时由于1 和 2,3是在两个线程里执行, 所以当1执行后, bFull已经等于true,
而此时2与3都可以执行, 至于谁先执行,那是不可预测的.有可能某个线程被由于系统资源或其它的原因被阻塞,因而导致延迟.

Thread.sleep(1);//加这句只是让get()执行的时间延长,那么3的执行时间就会相应延长.所以2就有更多的机会在3之前执行.

[lei198203]

由于本人表达能力不佳,所以不要见笑.

[lei198203]

改正错误:

但时由于1 和 2,3是在两个线程里执行
应该是
1,2 和 3 是在两个线程里执行(Producer 和 Consumer), 它们在操作系统里是并行的,没有次序.
1执行完之后, 3里的q.get()就可以返回结果, 因此3就可以将字串打印出来.