是否在螺纹完成时发出通知？为什么此代码示例工作？

问题描述

我在寻找某些谜题的线程，我不知道为什么下面一贯打印 999999 ：

I am looking in some puzzles for threads and I can't figure out why the following consistently prints 999999:

class Job extends Thread {
    private Integer number = 0;
    public void run() {
        for (int i = 1; i < 1000000; i++) {
            number++;
        }
    }
    public Integer getNumber() {
        return number;
    }
}
public class Test {
    public static void main(String[] args)
    throws InterruptedException {
        Job thread = new Job();
        thread.start();
        synchronized (thread) {
            thread.wait();
        }
        System.out.println(thread.getNumber());
    }
}

没有 / code>在同一个锁（和虚假的唤醒似乎被忽略）。

如果一个线程完成了一个通知得到信号或某事？

如何来到<$

There is no notify on the same lock (and spurious wakeup seem to be ignored).
If a thread finishes does a notify get signalled or something?
How come main prints the result and not get "stuck" waiting?

推荐答案

在Javadoc for Java 7

In the Javadoc for Java 7 Thread.join(long)

使用Thread直接这种方式被认为是不实际的。

Using a Thread directly this way is considered bad practical. Note: wait() could end for any number of reasons, possibly spuriously.

基于与@ Voo相关的解谜器评论。关键是你不应该使用Thread的内部行为，因为这更有可能导致混乱。

Based on a puzzler related to @Voo's comment. The point is you shouldn't play with the internal behaviour of Thread as this is more likely to lead to confusion.

public static String getName() {
    return "MyProgram";
}
public static void main(String... args) {
    new Thread() {
       public void run() {
           System.out.println("My program is " + getName());
        }
    }.start();
}

此程式列印什么？

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