On Wed, Sep 14, 2022 at 1:51 PM r r <grosi...@gmail.com> wrote:
> Hello,
> let's look for the following piece of code:
>
> int x;
> volatile boolean v; // v = false by default
> T1:
> x = 1; (1)
> v = true; (2)
> doSth (3)
>
> T2:
> doSth2 (4)
> if (v) {} (5)
>
>
> When T2 observes that v == false in (5), does it mean that there is a
> happens-before relation (4) -> (5) -> (2) -> (3)?
>
No.
There can't be a happens-before edge between a read of v (5) and a write of
v (4).
The volatile write/read will be ordered in the synchronization order, but
not in the synchronized-with order and therefore not ordered by
happens-before (since the happens-before order is the transitive closure
of the union of the synchronizes-with order and the program order).
Only when a volatile read sees a particular volatile write, then there is a
happens-before edge from the write to the read, but never in the opposite
direction.
>
> What if v would be AtomicBolean?
>
Doesn't change anything since an AtomicBoolean get/set has the same
semantics as a volatile read/write.
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