java线程可见性 [英] java thread visibility

问题描述

当我阅读实践中的java并发c03时，我对以下程序感到困惑：

When I read the "java concurrency in practice" c03, I was confused by the following program:

public class NoVisibility { 
    private static boolean ready; 
    private static int number; 

    private static class ReaderThread extends Thread { 
        public void run() { 
            while (!ready) 
                Thread.yield(); 
            System.out.println(number); 
        } 
    } 

    public static void main(String[] args) { 
        new ReaderThread().start(); 
        number = 42; 
        ready = true; 
    } 
}

由于重新排序和线程可见性，永远不会停止，或者输出可能为零，但我已经尝试过很多次，输出总是42.
所有的原因是我太幸运了吗？

Because of the reordering and thread visibility, the loop may never stop, or the output may be zero, but I have tried many times, and the output is always 42. All the reason is I'm too lucky?

推荐答案

所有原因是我太幸运了吗？

All the reason is I'm too lucky?

不一定。这将取决于你的处理器架构和JVM实现。这是微妙的记忆模型问题的问题之一：它们可能很难在野外繁殖。

Not necessarily. It will depend on your processor architecture and JVM implementation too. That's one of the problems with subtle memory model issues: they can be very hard to reproduce in the wild.

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