On Mar 17, 2004, at 7:39 PM, Thomas Wang wrote:
> Initially, x = y = v = 0. v is a volatile variable.
>
> Thread 1:
> r1 = x
> v = 0
> y = 1
>
> Thread 2:
> r3 = y
> v = 0
> x = 1
>
> Is the behavior r1 == r3 == 1 possible?
>
Yes (under both weak and strong interpretations)
As David noted, a write to a volatile serves as a release, and
"publishes" the memory accesses
that occur before it to anyone who later reads that volatile (with
slight differences for the strong/weak
interpretation).
For example, compilers can reorder a volatile write and a following
non-volatile memory access, resulting in:
thread 1:
y = 1
r1 = x
v = 0
thread 2:
x = 1
r3 = y
v = 0
-------------------------------
JavaMemoryModel mailing list - http://www.cs.umd.edu/~pugh/java/memoryModel
This archive was generated by hypermail 2b29 : Thu Oct 13 2005 - 07:01:00 EDT