On Thu, Sep 15, 2022 at 11:43 AM r r <[email protected]> wrote:
> What about a such case:
>
> AtomicLong x;
> volatile boolean v;
>
> T1:
> v = true; (1)
> long a = x.incrementAndGet(); // a == 1 (2)
> T2:
> long b = x.incrementAndGet(); // b == 2 (3)
>
> Do I understand correctly that if T2 observes that x == 2 it also means
> that T2 observer v == true because of happens-before *(1) -hb-> (2) -hb->
> (3)*?
>
'yes'.
So imagine 'v' would be plain and you would have the following code:
T1:
v=true (1)
x.incrementAndGet(); (2)
T2:
if(x.incrementAndGet()==2){ (3)
print(v) (4)
}
If (3) observes (2), then there is a happens-before edge between (1) and
(4) because:
(1) happens-before (2) due to program order rule
(2) happens-before (3) due to volatile variable rule
(3) happens-before (4) due to program order rule
Since happens-before is transitive, there is a happens-before edge between
(1) and (4).
So since we have a happens-before edge when v is plain, we certainly have a
happens-before edge is v would be volatile.
>
> środa, 14 września 2022 o 14:04:16 UTC+2 r r napisał(a):
>
>> Thanks
>>
>> śr., 14 wrz 2022, 13:26 użytkownik Peter Veentjer <[email protected]>
>> napisał:
>>
>>>
>>>
>>> On Wed, Sep 14, 2022 at 1:51 PM r r <[email protected]> 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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