Deep understanding of volatile in Java

问题

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?

回答1:

Any actions before a volatile write happen before (HB) any subsequent volatile read of the same variable. In your case, the write to x happens before the write to g (due to program order).

So there are only three possibilities:

• actor2 runs first and x and g are 0 - output is 0,0
• actor1 runs first and x and g are 1 because of the happens before relationship HB - output is 1,1
• the methods run concurrently and only x=1 is executed (not g=1) and the output could be either 0,1 or 0,0 (no volatile write so no guarantee)

回答2:

No, this isn't possible. According to the JMM, anything that was visible to thread 1 when it writes to a volatile field becomes visible to thread 2 when it reads that field.

There is another example similar to yours provided here:

class VolatileExample {
  int x = 0;
  volatile boolean v = false;
  public void writer() {
    x = 42;
    v = true;
  }

  public void reader() {
    if (v == true) {
      //uses x - guaranteed to see 42.
    }
  }
}

回答3:

No, and in fact this property of volatile is used in classes like ConcurrentHashMap to implement a lock-free happy path, roughly like this:

volatile int locked = 0;
...
void mutate() {
    if (Unsafe.compareAndSwapInt(locked,0,1)) { 
    /*this isn't exactly how you call this method, but the point stands: 
      if we read 0, we atomically replace it with 1 and continue on the happy 
      path */
       //we are happy
       //so we mutate the structure and then
       locked = 0;           
    } else {
       //contended lock, we aren't happy
    }
}

Since writes before a volatile write can't be reordered after the volatile write, and reads after volatile read can't be reordered before the volatile read, code like this indeed works as a "lockless locking".

来源：https://stackoverflow.com/questions/45133832/deep-understanding-of-volatile-in-java