System.out.format如何防止死锁?

Bob*_*ton 1 java concurrency deadlock

我发现在经典Java死锁教程中包含对System.out.format的调用可以防止发生死锁,我无法弄清楚原因.

下面的代码与教程相同,但增加mainSystem.out.format("Hi, I'm %s...no deadlock for you!\n\n", alphonse.getName());

public class Deadlock {
    static class Friend {
        private final String name;

        public Friend(String name) {
            this.name = name;
        }

        public String getName() {
            return this.name;
        }

        public synchronized void bow(Friend bower) {
            System.out.format("%s: %s has bowed to me!\n",
                    this.name, bower.getName());
            bower.bowBack(this);
        }

        public synchronized void bowBack(Friend bower) {
            System.out.format("%s: %s has bowed back to me!\n",
                    this.name, bower.getName());
        }
    }

    public static void main(String[] args) throws InterruptedException {
        final Friend alphonse = new Friend("Alphonse");
        final Friend gaston = new Friend("Gaston");

        System.out.format("Hi, I'm %s...no deadlock for you!\n\n", alphonse.getName());

        new Thread(new Runnable() {
            public void run() { alphonse.bow(gaston); }
        }).start();

        new Thread(new Runnable() {
            public void run() { gaston.bow(alphonse); }
        }).start();
    }
}
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这是输出:

Hi, I'm Alphonse...no deadlock for you!

Alphonse: Gaston has bowed to me!
Gaston: Alphonse has bowed back to me!
Gaston: Alphonse has bowed to me!
Alphonse: Gaston has bowed back to me!
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删除违规行会导致通常的死锁:

Alphonse: Gaston has bowed to me!
Gaston: Alphonse has bowed to me!
... deadlock ...
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对System.out.format的调用是否以某种方式改变了线程获取对象的内部锁的方式?

更新:

我只是通过改变我在代码中启动线程的位置来使系统再次死锁:

public static void main(String[] args) throws InterruptedException {
    final Friend alphonse = new Friend("Alphonse");
    final Friend gaston = new Friend("Gaston");

    System.out.format("Hi, I'm %s...no deadlock for you!\n\n", alphonse.getName());

    Thread t1 = new Thread(new Runnable() {
        public void run() { alphonse.bow(gaston); }
    });

    Thread t2 = new Thread(new Runnable() {
        public void run() { gaston.bow(alphonse); }
    });

    t1.start();
    t2.start();
}
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这引出了一个问题,即我们如何能够更深入地了解线程调度程序的行为,但我会在不同的日子保存它.

Op *_*kel 5

你没有真正删除死锁,而是(因为一些内部JVM原因)改变了线程的时间,以便其中一个线程bowBack() 其他调用之前进入bow().刚刚投入bow:sleep(1000)你的僵局将再次出现.

需要注意的是死锁不会总是发生,只有当线程都在幸运的时刻.在这种情况下,当两个线程都进入bow并且在它们中的任何一个调用之前,将发生死锁bowBack

...而"一些内部JVM原因"可能如下:

在你的情况下,实际上有三个线程:一个执行main,t1t2.将打印隐藏死锁的原因可能是线程调度程序决定main仍然有工作要做,即刷新io缓冲区,因此在启动t1之后和启动t2之前让main继续.如果你是一个双核CPU,只和会运行,但会等待,因为是一个缓慢的操作.上下文切换需要更多时间,并且t1将在t2开始之前完成...因此不会发生死锁.但这并不意味着如果你再次运行程序就不会发生死锁.maint1t2print

如果你想玩,queue在那个队列中创建一个并推送令牌(线程名称),然后join在main中创建你的线程.完成后,打印队列内容,您可以观察线程的时间.