Re: Two more multithreading questions

From:
Knute Johnson <nospam@rabbitbrush.frazmtn.com>
Newsgroups:
comp.lang.java.programmer
Date:
Wed, 31 Jan 2007 13:23:16 -0800
Message-ID:
<6N7wh.154015$cv2.102216@newsfe13.lga>
A. Bolmarcich wrote:

On 2007-01-31, Knute Johnson <nospam@rabbitbrush.frazmtn.com> wrote:

Thanks very much for your response. The two actions are independent and
I do not want to effect when they occur. I just want to ensure that if
an assignment is made to the variable that any subsequent read in the
other thread will have the latest value.


Without any other sychronization action between the threads, the only
way you know that a read was subsequent to a write is based on the value
that was read. A read is subsequent to a write that wrote the value
that was read.

The fact that the variable is volatile means that reads and writes by
a thread cannot be reordered to be before the previous synchronization
action or after the next synchronization action. According to section
"8.3.1.4 volatile Fields" of the JLS (from
http://java.sun.com/docs/books/jls/third_edition/html/classes.html#36930),
given the class

  class Test {
    static volatile int i = 0, j = 0;
    static void one() { i++; j++; }
    static void two() {
      System.out.println("i=" + i + " j=" + j);
    }
  }

If method one() is repeatedly called by one thread and method two() is
repeatedly called by another thread, then according to the JLS:

  Therefore, the shared value for j is never greater than that for i,
  because each update to i must be reflected in the shared value for i
  before the update to j occurs.

If the variables i and j were not volatile, then lines printed by method
two() may have a value of j greater than that of i.


So that's a yes?

--

Knute Johnson
email s/nospam/knute/

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