使用System.out.format和System.out.println进行多线程处理

Question

我在Oracle的Java Tutorial中遇到了这个例子,描述了多线程场景中的死锁.

所以在这个例子中,我在第17行和第18行进行了以下更改.

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) {
        //My Changes
        //System.out.format("%s: %s" + " has bowed to me!%n", this.name, bower.getName()); //Line 17
         System.out.println(this.name + ": " + bower.getName() + " has bowed to me!"); //Line 18
        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) {
    final Friend alphonse = new Friend("Alphonse");
    final Friend gaston = new Friend("Gaston");
    new Thread(new Runnable() {
        @Override
        public void run() {
            alphonse.bow(gaston);
        }
    }).start();

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

执行这些更改后,程序成功终止,不会导致死锁并打印后输出

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!

所以我的问题是 - 为什么它会像这样？println语句是如何防止死锁的？

Answer 1

你是否使用System.out.print或没有区别System.out.format:他们基本上做同样的事情.

如果Gaston.bow(Alphonse)在开始Alphonse.bow(Gaston)和之间开始执行bower.bowBack(Alphonse)(反之亦然),则发生死锁:两个线程正在等待另一个线程持有的监视器,因此发生死锁.

出现这种情况不一致,因为它依赖于一个微妙的时机的问题,这取决于线程是如何安排-这是可能的,Alphonse.bow并且bower.backBack(Alphonse)完成前Gaston.bow被执行,所以看起来没有僵局.

解决此问题的经典方法是订购锁定获取,以便每次首先获取相同的锁定; 这可以防止死锁的可能性:

public void bow(Friend bower) {  // Method no longer synchronized.
  int firstHash = System.identityHashCode(this);
  int secondHash = System.identityHashCode(bower);

  Object firstMonitor = firstHash < secondHash ? this : bower;
  Object secondMonitor = firstHash < secondHash ? bower : this;
  synchronized (firstMonitor) {
    synchronized (secondMonitor) {
      // Code free (*) of deadlocks, with respect to this and bower at least.
    }
  }
}

(*)它不能完全保证死锁,因为System.identityHashCode可以为不同的对象返回相同的值; 但这是不太可能的.

这是生日悖论的应用:如果你只有两台显示器,那么碰撞的可能性就像10 ^ -18; 但是如果你有超过77k的显示器,那么碰撞就更有可能了.