Hi_Amos
坚持每天都在进步!!

1.什么是Semaphore?

A counting semaphore. Conceptually, a semaphore maintains a set of permits. Each acquire blocks if 
 necessary until a permit is available, and then takes it. Each release adds a permit, potentially releasing a 
 blocking acquirer. However, no actual permit objects are used; the Semaphore just keeps a count of the 
 number available and acts accordingly. 
Semaphores are often used to restrict the number of threads than can access some (physical or logical) 
 resource.

上面是官方给予该类的介绍,信号量(Semaphore),有时被称为信号灯,是在多线程环境下使用的一种设施, 它负责协调各个线程, 以保证它们能够正确、合理的使用公共资源。

举个例子,公共厕所只有3个位子,有10个人要上厕所.一开始3个位都是空的,那么将有3个人先后占得坑,其它人只要在外面等待.如果有一个人上完厕所,然后等待中的7人中将有一个可以去上厕所.也有可能同时有两人OK,那么也将会同时有两人补位.简单说就是你必须有空位了,才能去上厕所.

上厕所的例子虽不雅观,但令人印象深刻,本来想举ATM机取钱的例子的,但意思一样,还是举上厕所吧.

Semaphore的作用就是控制位置的分配,一般情况下位置的分配是随机的,可以在实例化对象时设置规则进行排序.

2.实例

先看效果:

 

 

再来看代码:

 

package com.amos.concurrent;

import java.util.Random;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;

/** 
* @ClassName: SemaPhoreTest 
* @Description: 线程通信中的"信号灯"
* @author: amosli
* @email:hi_amos@outlook.com
* @date Apr 25, 2014 12:06:22 AM  
*/
public class SemaPhoreTest {
    public static void main(String[] args) {
        ExecutorService threadPool = Executors.newCachedThreadPool();
        
        final Semaphore semaphore=new Semaphore(3);
        for(int i=0;i<10;i++){
            threadPool.execute(new Runnable() {
                public void run() {
                    try {
                        semaphore.acquire();//获取一个可用的permits
                    } catch (Exception e) {
                        e.printStackTrace();
                    }
                    System.out.println("线程 " + Thread.currentThread().getName()+" 已进入.  " + "目前已经有"+(3-semaphore.availablePermits())+" 个线程进入");
                    try {
                        Thread.sleep(new Random().nextInt(1000));
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    System.out.println("线程 "+Thread.currentThread().getName()+ " 即将离开...");
                    semaphore.release();//释放一个线程
                    System.out.println("线程 "+Thread.currentThread().getName()+ " 已离开.  当前有"+(3-semaphore.availablePermits())+"并发");
                }
            });
        }
    }
}

 

怎么保证是有序的?

上面的代码是不保证顺序的,如果要有个先后顺序,即FIFO,先进先出原则,有个排队,上厕所要排队,只需要将上面的代码改掉一行即可.

final Semaphore semaphore=new Semaphore(3, true);

这样就可以了,其它代码不变.效果如下图所示:

 

3.代码说明

1).如何创建Semaphore?

new Semaphore(3);//无序默认创建是无序的.
new Semaphore(3,true);//有序

2)如何获取一个"坑位"?

首先,程序会先检查,"坑"是否已经满了,如果没满,那么可以按需分配,具体代码如下:

          try {semaphore.acquire();//获取一个可用的permits
                     } catch (Exception e) {
                        e.printStackTrace();
                    }

然后,分配完以后,将会把此坑位置为不可用了.

3)如何释放一个"坑位"?

semaphore.release();//释放一个线程

释放完成以后将会把此坑位置为可用的.

 

4.扩展--官方例子

For example, here is a class that uses a semaphore to control access to a pool of items: 
 class Pool {
   private static final int MAX_AVAILABLE = 100;
   private final Semaphore available = new Semaphore(MAX_AVAILABLE, true);

   public Object getItem() throws InterruptedException {
     available.acquire();
     return getNextAvailableItem();
   }

   public void putItem(Object x) {
     if (markAsUnused(x))
       available.release();
   }

   // Not a particularly efficient data structure; just for demo

   protected Object[] items = ... whatever kinds of items being managed
   protected boolean[] used = new boolean[MAX_AVAILABLE];

   protected synchronized Object getNextAvailableItem() {
     for (int i = 0; i < max_available; ++i) {
       if (!used[i]) {
          used[i] = true;
          return items[i];
       }
     }
     return null; // not reached
   }

   protected synchronized boolean markasunused(object item) {
     for (int i = 0; i < max_available; ++i) {
       if (item == items[i]) {
          if (used[i]) {
            used[i] = false;
            return true;
          } else
            return false;
       }
     }
     return false;
   }

 }

 

 

posted on 2014-04-25 01:09  Hi_Amos  阅读(1206)  评论(0编辑  收藏  举报