java并发编程学习:用 Semaphore (信号量)控制并发资源

简介: 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...

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

注意红色的部分:Thread0与Thread14同时分配到了Printer0上了,出现了多个线程同时抢一个资源的情况。

 


参考文章:

http://ifeve.com/thread-synchronization-utilities-2/

http://ifeve.com/thread-synchronization-utilities-3/

http://ifeve.com/concurrency-semaphore/

目录
相关文章
|
2月前
|
存储 Oracle Java
java零基础学习者入门课程
本课程为Java零基础入门教程,涵盖环境搭建、变量、运算符、条件循环、数组及面向对象基础,每讲配示例代码与实践建议,助你循序渐进掌握核心知识,轻松迈入Java编程世界。
319 0
|
2月前
|
Java 大数据 Go
从混沌到秩序:Java共享内存模型如何通过显式约束驯服并发?
并发编程旨在混乱中建立秩序。本文对比Java共享内存模型与Golang消息传递模型,剖析显式同步与隐式因果的哲学差异,揭示happens-before等机制如何保障内存可见性与数据一致性,展现两大范式的深层分野。(238字)
97 4
|
2月前
|
负载均衡 Java API
grpc-java 架构学习指南
本指南系统解析 grpc-java 架构,涵盖分层设计、核心流程与源码结构,结合实战路径与调试技巧,助你从入门到精通,掌握高性能 RPC 开发精髓。
324 7
|
2月前
|
IDE Java 编译器
java编程最基础学习
Java入门需掌握:环境搭建、基础语法、面向对象、数组集合与异常处理。通过实践编写简单程序,逐步深入学习,打牢编程基础。
236 1
|
2月前
|
缓存 安全 Java
如何理解Java中的并发?
Java并发指多任务交替执行,提升资源利用率与响应速度。通过线程实现,涉及线程安全、可见性、原子性等问题,需用synchronized、volatile、线程池及并发工具类解决,是高并发系统开发的关键基础。(238字)
250 5
|
2月前
|
Java
如何在Java中进行多线程编程
Java多线程编程常用方式包括:继承Thread类、实现Runnable接口、Callable接口(可返回结果)及使用线程池。推荐线程池以提升性能,避免频繁创建线程。结合同步与通信机制,可有效管理并发任务。
183 6
|
2月前
|
安全 前端开发 Java
从反射到方法句柄:深入探索Java动态编程的终极解决方案
从反射到方法句柄,Java 动态编程不断演进。方法句柄以强类型、低开销、易优化的特性,解决反射性能差、类型弱、安全性低等问题,结合 `invokedynamic` 成为支撑 Lambda 与动态语言的终极方案。
169 0
|
NoSQL Java Redis
Java程序中如何正确的释放资源(备战2022春招或暑期实习,每天进步一点点,打卡100天,Day2)
Java程序中如何正确的释放资源(备战2022春招或暑期实习,每天进步一点点,打卡100天,Day2)
189 0
Java程序中如何正确的释放资源(备战2022春招或暑期实习,每天进步一点点,打卡100天,Day2)
|
Java 程序员 API
Java程序中如何正确的释放资源
在Java开发中,程序员经常会遇到各种资源的释放问题。比如最常见的I/O操作,我们往往会通过调用API提供的close方法来关闭流,释放资源。但是追求极致的程序员会发现,这种方式存在不少问题,比如忘记关闭流、代码不美观、异常不好处理等等。
835 0
Java程序中如何正确的释放资源
|
2月前
|
JSON 网络协议 安全
【Java】(10)进程与线程的关系、Tread类;讲解基本线程安全、网络编程内容;JSON序列化与反序列化
几乎所有的操作系统都支持进程的概念,进程是处于运行过程中的程序,并且具有一定的独立功能,进程是系统进行资源分配和调度的一个独立单位一般而言,进程包含如下三个特征。独立性动态性并发性。
198 1