Re: Two more multithreading questions
On Jan 31, 4:24 pm, "A. Bolmarcich" <agge...@earl-grey.cloud9.net>
wrote:
On 2007-01-31, Daniel Pitts <googlegrou...@coloraura.com> wrote:
On Jan 31, 11:40 am, "A. Bolmarcich" <agge...@earl-grey.cloud9.net>
wrote:
[snip]
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 (fromhttp://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.
Actually, they still might.
imagine this scenario:
Thread 2: two gets called
Thread 2: value i is loaded and converted to a string for
concatication
Thread 1: one gets called
Thread 1: one gets called again
Thread 1: one gets called a third time
Thread 2: continues with j
Output is i=0 j=3
Which is a case of the value of j being greater than that of i that I wrote
may occur.
You wrote:
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.
My example is allowing them to be volatile.
Actually, my example is valid for both volatile and non-volatile
variables.
With two threads repeatedly calling their respective methods, you can
get a situation with i > j or j > i, regardless of the volatility of
the variables.
Therefore, your post doesn't explain volatile at all.
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