Re: Mutable Objects and Thread Boundaries

From:

Lew <noone@lewscanon.com>

Newsgroups:

comp.lang.java.programmer

Date:

Wed, 21 Jul 2010 09:46:50 -0400

Message-ID:

<i26to6$f8l$1@news.albasani.net>

Peter Duniho wrote:

class Test
{
private volatile boolean set = false;

In the above example, if methodA() is executed on one thread, and then
methodB() is executed in a different thread, methodB() is guaranteed
to see the new value for "data".

markspace wrote:

Yes, for volatile, this works. Volatile fields have special semantics
which prevent re-ordering reads and writes by the compiler, so this is
guaranteed to work.

I said it didn't work for all of the cases Joshua listed, particularily
the synchronized one. Replace the volatile with a synchronized block and
it won't work anymore.

It will so, as long as they synchronize on the same lock or monitor.

Just trying to let other folks know what works and what doesn't. I think
you knew this already, just that you didn't touch all the bases when you
mentioned it originally.

class Test
{
  private boolean set;
  private int data;

  void methodA()
  {
    data = 5;
    synchronized( this ) {
       set = true;
    }
  }

  void methodB()
  {
    synchronized( this ) {
      if (set)
      {
        System.out.print(data); // may print "0"
      }
    }
  }
}

You're mistaken. The assignment in 'methodA()' of 'data' /happens-before/ the
write to 'set', which latter is synchronized. The read in 'methodB()' of
'data' /happens-after/ the read of 'set', which latter is synchronized on the
same monitor. Therefore the read is guaranteed to see the write.

--
Lew