java并发编程学习：用 Semaphore （信号量）控制并发资源

并发编程这方面以前关注得比较少，恶补一下，推荐一个好的网站：并发编程网 – ifeve.com，上面全是各种大牛原创或编译的并发编程文章。

今天先来学习Semaphore（信号量），字面上看，根本不知道这东西是干啥的，借用并发工具类（三）控制并发线程数的Semaphore一文中的交通红绿信号灯的例子来理解一下：

一条4车道的主干道，假设100米长，每辆车假设占用的长度为10米(考虑到前后车距)，也就是说这条道上满负载运行的话，最多只能容纳4*(100/10)=40辆车，如果有120辆车要通过的话（为简单起见，一波40辆，分成3波），就必须要红绿信号灯来调度了，对于最前面的一波来讲，它们看到的是绿灯，允许通过，第一波全进入道路后，红绿灯变成红色，表示后面的2波，要停下来等候第1波车辆全通过，然后红绿灯才会变成绿色，让第2波通过，如此运转下去….

这跟多线程并发有啥关系呢？Semaphore就是红绿信号灯，3波车辆就是3个并发的线程，而主干道就是多个线程要并发访问的公用资源，由于资源有限，所以必须通过Semaphore来控制线程对资源的访问，否则就变成资源竞争，严重的话会导致死锁等问题。

下面用一个示例演示，假设有N个并发线程都要打印文件，但是打印机只有1台，先来一个打印队列类：

package yjmyzz.lesson01;import java.util.concurrent.Semaphore;public class PrintQueue {    private final Semaphore semaphore;    public PrintQueue() {
        semaphore = new Semaphore(1);//限定了共享资源只能有1个（相当于只有一把钥匙）
    }    public void printJob(Object document) {
        try {
            semaphore.acquire();//取得对共享资源的访问权（即拿到了钥匙））            long duration = (long) (1 + Math.random() * 10);
            System.out.printf("%s: PrintQueue: Printing a Job during %d seconds\n", Thread.currentThread().getName(), duration);
            Thread.sleep(duration);
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            semaphore.release();//钥匙用完了，要还回去，这样其它线程才能继续有序的拿到钥匙，访问资源
        }
    }
}

由于是在多线程环境中，真正运行的作业处理，得继承自Runnable(或Callable)

package yjmyzz.lesson01;public class Job implements Runnable {    private PrintQueue printQueue;    public Job(PrintQueue printQueue) {
        this.printQueue = printQueue;
    }    public void run() {
        System.out.printf("%s: Going to print a job\n", Thread.currentThread().getName());
        printQueue.printJob(new Object());
        System.out.printf("%s: The document has been printed\n", Thread.currentThread().getName());
    }
}

好了，测试一把：

package yjmyzz.lesson01;public class Main {
    public static void main(String args[]) {        PrintQueue printQueue = new PrintQueue();        int threadCount = 3;        Thread thread[] = new Thread[threadCount];
        for (int i = 0; i < threadCount; i++) {
            thread[i] = new Thread(new Job(printQueue), "Thread" + i);
        }        for (int i = 0; i < threadCount; i++) {
            thread[i].start();
        }
    }
}

输出：

Thread0: Going to print a job
Thread2: Going to print a job
Thread1: Going to print a job
Thread0: PrintQueue: Printing a Job during 7 seconds
Thread0: The document has been printed
Thread2: PrintQueue: Printing a Job during 5 seconds
Thread2: The document has been printed
Thread1: PrintQueue: Printing a Job during 1 seconds
Thread1: The document has been printed

从输出上看，线程0打印完成后，线程2才开始打印，然后才是线程1，没有出现一哄而上，抢占打印机的情况。这样可能没啥感觉，我们把PrintQueue如果去掉Semaphore的部分，变成下面这样：

package yjmyzz.lesson01;public class PrintQueue {    //private final Semaphore semaphore;    public PrintQueue() {
        //semaphore = new Semaphore(1);//限定了共享资源只能有1个（相当于只有一把钥匙）
    }    public void printJob(Object document) {
        try {
            //semaphore.acquire();//取得对共享资源的访问权（即拿到了钥匙））
            long duration = (long) (1 + Math.random() * 10);
            System.out.printf("%s: PrintQueue: Printing a Job during %d seconds\n", Thread.currentThread().getName(), duration);
            Thread.sleep(duration);
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            //semaphore.release();//钥匙用完了，要还回去，这样其它线程才能继续有序的拿到钥匙，访问资源
        }
    }
}

这回的输出：

Thread0: Going to print a job
Thread2: Going to print a job
Thread1: Going to print a job
Thread2: PrintQueue: Printing a Job during 4 seconds
Thread1: PrintQueue: Printing a Job during 8 seconds
Thread0: PrintQueue: Printing a Job during 1 seconds
Thread0: The document has been printed
Thread2: The document has been printed
Thread1: The document has been printed

可以发现，3个线程全都一拥而上，同时开始打印，也不管打印机是否空闲，实际应用中，这样必然出问题。

好的，继续，突然有一天，公司有钱了，又买了2台打印机，这样就有3台打印机了，这时候怎么办呢？简单的把PrintQueue构造器中的

    public PrintQueue() {
        semaphore = new Semaphore(3);
    }

就行了吗？仔细想想，就会发现问题，代码中并没有哪里能告诉线程哪个打印机正在打印，哪个打印机当前空闲，所以仍然有可能出现N个线程(N<=3)同时抢一台打印机的情况（即：如果把控制权当成钥匙的话，相当于有可能3个人各领取到了1把钥匙，但是这3把钥匙是相同的，3个人都看中了同一个箱子，都要用手中的钥匙去抢着开箱）。

所以得改进一下：

package yjmyzz.lesson02;import java.util.concurrent.Semaphore;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;public class PrintQueue {    private boolean freePrinters[];//用来存放打印机的状态，true表示空闲，false表示正在打印    private Lock lockPrinters;//增加了锁，保证多个线程，只能获取得锁，才能查询哪台打印机空闲的    private final Semaphore semaphore;    public PrintQueue() {
        int printerNum = 3;//假设有3台打印机
        semaphore = new Semaphore(printerNum);
        freePrinters = new boolean[printerNum];        for (int i = 0; i < printerNum; i++) {
            freePrinters[i] = true;//初始化时，默认所有打印机都空闲
        }
        lockPrinters = new ReentrantLock();
    }    private int getPrinter() {
        int ret = -1;
        try {
            lockPrinters.lock();//先加锁，保证1次只能有1个线程来获取空闲的打印机
            for (int i = 0; i < freePrinters.length; i++) {
                //遍历所有打印机的状态，发现有第1个空闲的打印机后，领取号码，
                // 并设置该打印机为繁忙状态（因为马上就要用它）
                if (freePrinters[i]) {
                    ret = i;
                    freePrinters[i] = false;
                    break;
                }
            }
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            //最后别忘记了解锁，这样后面的线程才能上来领号
            lockPrinters.unlock();
        }
        return ret;
    }    public void printJob(Object document) {
        try {
            semaphore.acquire();            int assignedPrinter = getPrinter();//领号
            long duration = (long) (1 + Math.random() * 10);
            System.out.printf("%s: PrintQueue: Printing a Job in Printer%d during %d seconds\n", Thread.currentThread().getName(),
                    assignedPrinter, duration);
            Thread.sleep(duration);
            freePrinters[assignedPrinter] = true;//打印完以后，将该打印机重新恢复为空闲状态        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            semaphore.release();
        }
    }
}

测试一下，这回把线程数增加到5，输出结果类似下面这样：

Thread0: Going to print a job
Thread4: Going to print a job
Thread3: Going to print a job
Thread2: Going to print a job
Thread1: Going to print a job
Thread4: PrintQueue: Printing a Job in Printer1 during 7 seconds
Thread0: PrintQueue: Printing a Job in Printer0 during 4 seconds
Thread3: PrintQueue: Printing a Job in Printer2 during 8 seconds
Thread0: The document has been printed
Thread2: PrintQueue: Printing a Job in Printer0 during 1 seconds
Thread2: The document has been printed
Thread4: The document has been printed
Thread1: PrintQueue: Printing a Job in Printer0 during 1 seconds
Thread3: The document has been printed
Thread1: The document has been printed

从输出结果可以看出，一次最多只能有3个线程使用这3台打印机，而且每个线程使用的打印机互不冲突，打印完成后，空闲的打印机会给其它线程继续使用，继续折腾，如果把getPrinter()中加锁的部分去掉，即：

    private int getPrinter() {
        int ret = -1;
        try {
            //lockPrinters.lock();//先加锁，保证1次只能有1个线程来获取空闲的打印机
            for (int i = 0; i < freePrinters.length; i++) {
                //遍历所有打印机的状态，发现有第1个空闲的打印机后，领取号码，
                // 并设置该打印机为繁忙状态（因为马上就要用它）
                if (freePrinters[i]) {
                    ret = i;
                    freePrinters[i] = false;
                    break;
                }
            }
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            //最后别忘记了解锁，这样后面的线程才能上来领号
            //lockPrinters.unlock();
        }
        return ret;
    }

再跑一下，结果如何，为了放大冲突，这回开到15个线程来抢3台打印机，输出如下：

Thread0: Going to print a job
Thread14: Going to print a job
Thread13: Going to print a job
Thread12: Going to print a job
Thread11: Going to print a job
Thread10: Going to print a job
Thread9: Going to print a job
Thread8: Going to print a job
Thread7: Going to print a job
Thread6: Going to print a job
Thread5: Going to print a job
Thread4: Going to print a job
Thread3: Going to print a job
Thread2: Going to print a job
Thread1: Going to print a job
Thread0: PrintQueue: Printing a Job in Printer0 during 29 seconds
Thread14: PrintQueue: Printing a Job in Printer0 during 92 seconds
Thread13: PrintQueue: Printing a Job in Printer1 during 66 seconds
Thread0: The document has been printed
Thread12: PrintQueue: Printing a Job in Printer0 during 86 seconds
Thread13: The document has been printed
Thread11: PrintQueue: Printing a Job in Printer1 during 1 seconds
Thread11: The document has been printed
Thread10: PrintQueue: Printing a Job in Printer1 during 58 seconds
Thread14: The document has been printed
Thread9: PrintQueue: Printing a Job in Printer0 during 92 seconds
Thread12: The document has been printed
Thread8: PrintQueue: Printing a Job in Printer0 during 59 seconds
Thread10: The document has been printed
Thread7: PrintQueue: Printing a Job in Printer1 during 51 seconds
Thread8: The document has been printed
Thread6: PrintQueue: Printing a Job in Printer0 during 33 seconds
Thread7: The document has been printed
Thread5: PrintQueue: Printing a Job in Printer1 during 2 seconds
Thread9: The document has been printed
Thread3: PrintQueue: Printing a Job in Printer1 during 85 seconds
Thread4: PrintQueue: Printing a Job in Printer0 during 61 seconds
Thread5: The document has been printed
Thread6: The document has been printed
Thread2: PrintQueue: Printing a Job in Printer0 during 66 seconds
Thread4: The document has been printed
Thread1: PrintQueue: Printing a Job in Printer0 during 9 seconds
Thread1: The document has been printed
Thread3: The document has been printed
Thread2: The document has been printed