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java thread visibility

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; 
    } 
}

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?

like image 465
xing.zhang Avatar asked Mar 26 '12 07:03

xing.zhang


1 Answers

All the reason is I'm too lucky?

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.

like image 174
Jon Skeet Avatar answered Oct 23 '22 07:10

Jon Skeet