对于此实现，如何确保打印偶数奇数的两个线程先保持偶数，然后再保持奇数？

Question

我创建了两个可运行的作业：PrintEvenNumbersJob 和 PrintOddNumbersJob，并生成了两个线程来执行这些作业。这似乎工作得很好！但是我闻到了这个实现的可疑之处。我可以就这个实现提供一些 cmets 和建议吗？

我在这个实现中看到的问题是，程序只有在 thread1 首先获得对象锁的锁时才会终止，否则它会打印奇数的第一个偶数个第二个顺序，并且除非我提供另一个语句，否则它不会终止PrintEvenNumbersJob 中的 for 语句之后的“lock.notify”（如在此实现中）。我的问题是如何确保首先执行thread1。

public class PrintEvenNumbersJob implements Runnable {

private Object lock;

public PrintEvenNumbersJob(Object lock) {
    this.lock = lock;
}

@Override
public void run() {
    synchronized (lock) {
        for (int i = 0; i <= 10; i += 2) {

            lock.notify();

            System.out.println(i);

            try {
                lock.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }
        lock.notify(); // not required if thread1 gains lock first
    }
}

}


public class PrintOddNumbersJob implements Runnable {

private Object lock;

public PrintOddNumbersJob(Object lock) {
    this.lock = lock;
}

@Override
public void run() {
    synchronized (lock) {
        for (int i = 1; i < 10; i += 2) {

            lock.notify();

            System.out.println(i);

            try {
                lock.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }
        lock.notify();
    }
}

}



public class EvenOddManager {

public static void main(String[] args) {

    Object lock = new Object();

    PrintEvenNumbersJob printEvenNumbersJob = new PrintEvenNumbersJob(lock);
    PrintOddNumbersJob printOddNumbersJob = new PrintOddNumbersJob(lock);

    Thread thread1 = new Thread(printEvenNumbersJob);
    Thread thread2 = new Thread(printOddNumbersJob);

    thread2.start();
    thread1.start();

}

}

Answer 1

您是否尝试过使用信号量？这更容易，因为您无需担心调用 wait 和 notify 的顺序（如果您在等待之前调用 notify，它会“丢失”）

示例代码：

import java.util.concurrent.*;

public class Test {

    private final Semaphore oddJobPermits = new Semaphore(0);
    private final Semaphore evenJobPermits = new Semaphore(1);

    private class EvenJob implements Runnable {
        public void run() {
            for (int i = 0; i < 10; i++) {
                try {
                    evenJobPermits.acquire();
                    System.out.println(i * 2);
                } catch (InterruptedException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                } finally {
                    oddJobPermits.release();
                }
            }
        }
    }

    private class OddJob implements Runnable {
        public void run() {
            for (int i = 0; i < 10; i++) {
                try {
                    oddJobPermits.acquire();
                    System.out.println(i * 2 + 1);
                } catch (InterruptedException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                } finally {
                    evenJobPermits.release();
                }
            }
        }
    }

    public void run() {
        new Thread(new EvenJob()).start();
        new Thread(new OddJob()).start();
    }

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

}

Answer 2

我相信你需要一个裁判：

public class Referee {
    private boolean evensTurn = true;

    public void waitMyTurn(boolean even) {
        synchronized(this) {
            while (even != evensTurn) {
                try {
                    wait();
                } finally {
                }
            }
        }
    }

    public void done() {
        synchronized(this) {
            evensTurn = !evensTurn;
            notify();
        }
    }
}

public class PrintEvenNumbersJob implements Runnable {

    private Referee referee;

    public PrintEvenNumbersJob(Referee referee) {
        this.referee = referee;
    }

    @Override
    public void run() {
        for (int i = 0; i <= 10; i += 2) {

            referee.waitMyTurn(true);

            System.out.println(i);

            referee.done();
        }
    }
}

public class PrintOddNumbersJob implements Runnable {

    private Referee referee;

    public PrintOddNumbersJob(Referee referee) {
        this.referee = referee;
    }

    @Override
    public void run() {
        for (int i = 0; i <= 10; i += 2) {

            referee.waitMyTurn(false);

            System.out.println(i);

            referee.done();
        }
    }
}

Answer 3

我尝试并测试了这段代码。它使用信号量工作

public class TestSemaphore
{

    public static void main(String[] args)
        throws Exception
    {
        AtomicInteger count = new AtomicInteger();
        Semaphore s = new Semaphore(1, true);
        Semaphore t = new Semaphore(1, true);

        OddNumberThread oThread = new OddNumberThread(count, s, t);
        EvenNumberThread eThread = new EvenNumberThread(count, s, t);

        eThread.start();
        oThread.start();
    }

    static class EvenNumberThread
        extends Thread
    {
        private AtomicInteger count;

        private Semaphore s, t;

        public EvenNumberThread(AtomicInteger pCount, Semaphore pS, Semaphore pT)
        {
            super("Even");
            count = pCount;
            s = pS;
            t = pT;
        }

        @Override
        public void run()
        {
            // Make this thread wait until even thread starts, Order will be incorrect if removed these lines.
            s.acquireUninterruptibly();
            while (count.intValue() <= 10)
            {
                try
                {
                    // Double checking to make it work
                    s.acquireUninterruptibly();

                    System.out.println(getName() + " " + count.getAndIncrement());
                }
                finally
                {
                    t.release();
                }
            }
        }
    }

    static class OddNumberThread
        extends Thread
    {
        private AtomicInteger count;

        private Semaphore s, t;

        public OddNumberThread(AtomicInteger pCount, Semaphore pS, Semaphore pT)
        {
            super("Odd");
            count = pCount;
            s = pS;
            t = pT;
        }

        @Override
        public void run()
        {
            // Start this thread first and start printing, Order will be incorrect if removed these lines.
            t.acquireUninterruptibly();
            s.release();

            while (count.intValue() <= 10)
            {
                try
                {
                    t.acquireUninterruptibly();

                    System.out.println(getName() + " " + count.getAndIncrement());
                }
                finally
                {
                    s.release();
                }
            }
        }
    }
}