java并发编程学习4--线程通信练习

【问题的描述：

启动3个线程打印递增的数字, 线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到打印到75. 程序的输出结果应该为:

线程1: 1
线程1: 2
线程1: 3
线程1: 4
线程1: 5

线程2: 6
线程2: 7
线程2: 8
线程2: 9
线程2: 10
...
线程3: 71
线程3: 72
线程3: 73
线程3: 74
线程3: 75

【代码实现

import java.util.ArrayList;
import java.util.List;

/**
 * Created by ibm on 2017/8/8.
 */
public class ClassicTest {

    //使用原始的synchornized object.wait() object.notify()
    public static void main(String[] args) {
        //定义线程组
        List<MyThread> threadGroups = new ArrayList<>();
        Counter counter = new Counter();
        MyThread t1 = new MyThread(1,"一",counter,threadGroups);
        MyThread t2 = new MyThread(2,"二",counter,threadGroups);
        MyThread t3 = new MyThread(2,"三",counter,threadGroups);
        threadGroups.add(t1);
        threadGroups.add(t2);
        threadGroups.add(t3);
        new Thread(t1).start();
        new Thread(t2).start();
        new Thread(t3).start();
    }
}

class MyThread implements Runnable{
    //线程运行状态 1马上执行，2阻塞
    public int status;
    //线程名字
    public String name;
    //计数器
    public Counter counter;
    //线程组
    public List<MyThread> threads = new ArrayList<>();

    public MyThread(int status,String name,Counter counter,List<MyThread> threads){
        this.status = status;
        this.name = name;
        this.counter = counter;
        this.threads = threads;
    }

    @Override
    public void run() {
        System.out.println(name + " GET " + status);
        synchronized (counter){
            while (!counter.isEnd()){
                //判断是否该自己执行，切记使用while,因为如果在循环的等待过程中status有所变化，这里需要再次判断
                while (status != 1){
                    try {
                        counter.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                for(int i = 0;i < 5 ;i ++){
                    System.out.println(name + ":" + counter.get());
                    counter.increase();
                }
                //状态进入阻塞状态，并设置下一个可以运行的线程
                status = 2;
                setNext();
                counter.notifyAll();
                System.out.println("----");
            }
        }
    }

    void setNext(){
        //当前线程在线程组的索引
        int index = 0;
        for(index = 0;index < threads.size();index++){
            if(name.equals(threads.get(index).name)){
                break;
            }
        }
        if(index == (threads.size() - 1)){
            threads.get(0).status = 1;
        }else {
            threads.get(index + 1).status = 1;
        }
    }
}

class Counter{

    int num = 1;
    int end = 75;

    public int get(){
        return num;
    }
    public void increase(){
        if(isEnd()){
            System.out.println("num超过限制");
            return;
        }
        num++;
    }
    public boolean isEnd(){
        if(num >= end){
            return true;
        }
        return false;
    }
}

【后面两种解法来源网络：

1.使用synchronized关键字：

public class ClassicTest1  {
    // n为即将打印的数字
    private static int n = 1;
    // state=1表示将由线程1打印数字, state=2表示将由线程2打印数字, state=3表示将由线程3打印数字
    private static int state = 1;

    public static void main(String[] args) {
        final ClassicTest1 pn = new ClassicTest1();
        new Thread(new Runnable() {
            public void run() {
                // 3个线程打印75个数字, 单个线程每次打印5个连续数字, 因此每个线程只需执行5次打印任务. 3*5*5=75
                for (int i = 0; i < 5; i++) {
                    // 3个线程都使用pn对象做锁, 以保证每个交替期间只有一个线程在打印
                    synchronized (pn) {
                        // 如果state!=1, 说明此时尚未轮到线程1打印, 线程1将调用pn的wait()方法, 直到下次被唤醒
                        while (state != 1)
                            try {
                                pn.wait();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        // 当state=1时, 轮到线程1打印5次数字
                        for (int j = 0; j < 5; j++) {
                            // 打印一次后n自增
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        // 线程1打印完成后, 将state赋值为2, 表示接下来将轮到线程2打印
                        state = 2;
                        // notifyAll()方法唤醒在pn上wait的线程2和线程3, 同时线程1将退出同步代码块, 释放pn锁.
                        // 因此3个线程将再次竞争pn锁
                        // 假如线程1或线程3竞争到资源, 由于state不为1或3, 线程1或线程3将很快再次wait, 释放出刚到手的pn锁.
                        // 只有线程2可以通过state判定, 所以线程2一定是执行下次打印任务的线程.
                        // 对于线程2来说, 获得锁的道路也许是曲折的, 但前途一定是光明的.
                        pn.notifyAll();
                    }
                }
            }
        }, "线程1").start();

        new Thread(new Runnable() {
            public void run() {
                for (int i = 0; i < 5; i++) {
                    synchronized (pn) {
                        while (state != 2)
                            try {
                                pn.wait();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        for (int j = 0; j < 5; j++) {
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        state = 3;
                        pn.notifyAll();
                    }
                }
            }
        }, "线程2").start();

        new Thread(new Runnable() {
            public void run() {
                for (int i = 0; i < 5; i++) {
                    synchronized (pn) {
                        while (state != 3)
                            try {
                                pn.wait();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        for (int j = 0; j < 5; j++) {
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        state = 1;
                        pn.notifyAll();
                    }
                }
            }
        }, "线程3").start();
    }
}

2.使用condition与lock

public class ClassicTest2 implements Runnable {
    private int state = 1;
    private int n = 1;
    // 使用lock做锁
    private ReentrantLock lock = new ReentrantLock();
    // 获得lock锁的3个分支条件
    private Condition c1 = lock.newCondition();
    private Condition c2 = lock.newCondition();
    private Condition c3 = lock.newCondition();

    @Override
    public void run() {
        new Thread(new Runnable() {
            public void run() {
                for (int i = 0; i < 5; i++) {
                    try {
                        // 线程1获得lock锁后, 其他线程将无法进入需要lock锁的代码块.
                        // 在lock.lock()和lock.unlock()之间的代码相当于使用了synchronized(lock){}
                        lock.lock();
                        while (state != 1)
                            try {
                                // 线程1竞争到了lock, 但是发现state不为1, 说明此时还未轮到线程1打印.
                                // 因此线程1将在c1上wait
                                // 与解法一不同的是, 三个线程并非在同一个对象上wait, 也不由同一个对象唤醒
                                c1.await();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        // 如果线程1竞争到了lock, 也通过了state判定, 将执行打印任务
                        for (int j = 0; j < 5; j++) {
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        // 打印完成后将state赋值为2, 表示下一次的打印任务将由线程2执行
                        state = 2;
                        // 唤醒在c2分支上wait的线程2
                        c2.signal();
                    } finally {
                        // 打印任务执行完成后需要确保锁被释放, 因此将释放锁的代码放在finally中
                        lock.unlock();
                    }
                }
            }
        }, "线程1").start();

        new Thread(new Runnable() {
            public void run() {
                for (int i = 0; i < 5; i++) {
                    try {
                        lock.lock();
                        while (state != 2)
                            try {
                                c2.await();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        for (int j = 0; j < 5; j++) {
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        state = 3;
                        c3.signal();
                    } finally {
                        lock.unlock();
                    }
                }
            }
        }, "线程2").start();

        new Thread(new Runnable() {
            public void run() {
                for (int i = 0; i < 5; i++) {
                    try {

                        lock.lock();
                        while (state != 3)
                            try {
                                c3.await();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        for (int j = 0; j < 5; j++) {
                            System.out.println(Thread.currentThread().getName()
                                    + ": " + n++);
                        }
                        System.out.println();
                        state = 1;
                        c1.signal();
                    } finally {
                        lock.unlock();
                    }
                }
            }
        }, "线程3").start();
    }

    public static void main(String[] args) {
        new ClassicTest2().run();
    }
}