深入理解Java中的volatile(Deep understanding of volatile in Java)

 Java允许输出1, 0吗？ 我已经对它进行了非常密集的测试，但我无法获得该输出。 我只得到1, 1或0, 0或0, 1 。 
 
public class Main {
    private int x;
    private volatile int g;

    // Executed by thread #1
    public void actor1(){
       x = 1;
       g = 1;
    }

    // Executed by thread #2
    public void actor2(){
       put_on_screen_without_sync(g);
       put_on_screen_without_sync(x);
    }
}
 
 为什么？ 
 
 在我眼中，有可能获得1, 0 。 我的理由。 g是易失性的，因此会确保内存顺序。 所以，它看起来像： 
 
actor1:

(1) store(x, 1)
(2) store(g, 1)
(3) memory_barrier // on x86
 
 并且，我看到以下情况：在store(x,1)之前重新排序store(g, 1) store(x,1) （memory_barrier 在 （2）之后）。 现在，运行线程＃2。 所以， g = 1, x = 0 。 现在，我们已经预期了产量。 我的推理有什么不对？ 

Does Java allows output 1, 0? I've tested it very intensively and I cannot get that output. I get only 1, 1 or 0, 0 or 0, 1.
 
public class Main {
    private int x;
    private volatile int g;

    // Executed by thread #1
    public void actor1(){
       x = 1;
       g = 1;
    }

    // Executed by thread #2
    public void actor2(){
       put_on_screen_without_sync(g);
       put_on_screen_without_sync(x);
    }
}
 
Why?
 
On my eye it is possible to get 1, 0. My reasoning. g is volatile so it causes that memory order will be ensured. So, it looks like:
 
actor1:

(1) store(x, 1)
(2) store(g, 1)
(3) memory_barrier // on x86
 
and, I see the following situation: reorder store(g, 1) before store(x,1) (memory_barrier is after (2)). Now, run thread #2. So, g = 1, x = 0. Now, we have expected output. What is incorrect in my reasoning? 

原文：https://stackoverflow.com/questions/45133832