如何确保两个线程打印奇数仍然保持这个实现的第一个奇数顺序？ [英] How to make sure two threads printing even odd numbers maintain even first then odd order for this implementation?

问题描述

我创建了两个可运行的作业：PrintEvenNumbersJob和PrintOddNumbersJob，并生成两个线程来执行这些作业。这似乎工作得很好！但我闻到了这个实施的可疑之处。我可以对此实现有一些意见和建议吗？

I have created two runnable jobs: PrintEvenNumbersJob and PrintOddNumbersJob and spawned two threads to execute these jobs. This seems to work perfectly fine! But I smell something suspicious about this implementation. Can I have some comments and advice on this implementation?

我看到这个实现的问题是程序只有在thread1获得锁定时才会终止对象锁定首先否则它打印奇数第一个偶数第二个订单并且不会终止，除非我在PrintEvenNumbersJob中的for语句之后提供另一个语句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();

}

}

推荐答案

您尝试使用信号量吗？它更容易，因为你不需要担心调用wait和notify的顺序（如果你在等待之前调用notify，它就丢失）

Have you try using Semaphores? It's easier because you don't need to worry about the order that wait and notify are called (if you call notify before the wait, it's "lost")

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();
    }

}

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