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:[email protected]
* @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;
} }