多线程解决生产者/消费者问题

weixin_30654583

于 2011-05-04 23:10:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/kaichd/archive/2011/05/04/2037083.html

本实验通过Java多线程技术实现生产者消费者问题的解决方案，利用信号量进行资源同步，确保生产者和消费者之间的正确交互。实验中定义了生产者、消费者及信号量类，模拟了生产消费过程。

这是本学期的操作系统课程的一个上机实验，用多线程来解决生产者消费者问题，源代码如下，我是初学者，尽量写的浅显些，尽量把注释写的完善些，如果有什么看不懂的可以留言，大家一起学习。

Semaphore.java

```
public class Semaphore {
```
```
    int value;
```

    public Semaphore(int v){this.value = v;}

```
    
```
```
    //定义P原语操作
```
```
    public synchronized void p(){
```
```
        value = value-1;
```
```
        if (value < 0){
```
```
            try{
```
```
                this.wait();
```
```
            }
```

            catch (InterruptedException e){

```
            }   
```
```
        }
```
```
    }
```
```
    
```
```
    //定义V原语操作
```
```
    public synchronized void v(){
```
```
        value = value+1;
```
```
        if (value <= 0){
```
```
            this.notify();
```
```
            }
```
```
        }
```
```
}
```
```
    
```

Producer.java

```
public class Producer extends Thread{
```
```
    private Semaphore[] s;
```
```
    private int i;
```

    public Producer(int i,Semaphore[] sp){

```
    this.i = i;
```
```
    this.s = sp;
```
```
    }
```
```
    public void run() {
```
```
        while(true){
```
```
            s[1].p();
```
```
            s[0].p();
```

            System.out.println("produce "+i++);

```
            try {
```
```
                Thread.sleep(30);
```
```
                }
```

            catch (InterruptedException e) {}

```
            s[0].v();
```
```
            s[2].v();
```
```
            }
```
```
        }
```
```
    }
```

Consumer.java

```
public class Consumer extends Thread {
```
```
    private Semaphore[] s;
```
```
    private int i;
```

    public Consumer(int i,Semaphore[] sp){

```
    this.i = i;
```
```
    this.s = sp;
```
```
    }
```
```
    public void run() {    
```
```
        while(true){
```
```
            s[2].p();
```
```
            s[0].p();
```

            System.out.println("consumer "+i++);

```
            try {
```

                Thread.sleep((long)(Math.random()*100));

```
                } 
```

            catch (InterruptedException e) {}

```
            s[0].v();
```
```
            s[1].v();
```
```
            }
```
```
        }
```
```
    }
```

pc.java

```
public class pc {
```

public static void main(String[] args) {

    int mutex = 1;//互斥的信号量，生产者消费者不能同时访问缓冲区

    int empty = 3;//表示空闲的缓冲区的数目，初始值为3

    int full = 0;//full表示有数据的缓冲区的数目，初始值为0

```
    Semaphore[] s = new Semaphore[3];
```
```
    s[0] = new Semaphore(mutex);
```
```
    s[1] = new Semaphore(empty);
```
```
    s[2] = new Semaphore(full);
```
```
    Producer p = new Producer(0,s);
```
```
    Consumer c = new Consumer(0,s);
```
```
    Thread pp = new Thread(p);
```
```
    Thread cc = new Thread(c);
```
```
    Thread pp1 = new Thread(p);
```
```
    Thread cc1 = new Thread(c);
```
```
    cc1.start();
```
```
    pp1.start();
```
```
    pp.start();
```
```
    cc.start();
```
```
    }
```
```
}
```

转载于:https://www.cnblogs.com/kaichd/archive/2011/05/04/2037083.html

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