[concurrency-interest] On A Formal Definition of 'Data-Race'

Gregg Wonderly gregg at cytetech.com
Wed Apr 17 09:25:21 EDT 2013


Actually the compiler doesn't do that for non-volatile access, as I've been 
going on about with the loop exit hoisting example.  This transformation is 
exactly the kind of thing that I'd expect to see happen, and this is the 
"unexplainable behavior" that will create alarm, because mutations of "shared" 
are not visible, except a loop exit.

Gregg Wonderly

On 4/17/2013 5:18 AM, oleksandr otenko wrote:
> You need to prove System.out.println isn't using shared.
>
> Alex
>
> On 17/04/2013 07:38, Nathan Reynolds wrote:
>> Couldn't JIT hoist the non-volatile writes out of the loop?  For example, the
>> following code...
>>
>> for (i = 0; i < 1_000_000_000; i++)
>> {
>>     System.out.println(i);
>>     shared = 2 * i;
>> }
>>
>> ... could be transformed into ...
>>
>> for (i = 0; i < 1_000_000_000; i++)
>> {
>>     System.out.println(i);
>> }
>>
>> shared = 2 * 1_000_000_000;
>>
>> ... If so, then the non-volatile write may not happen for a very long time.
>>
>> Nathan Reynolds <http://psr.us.oracle.com/wiki/index.php/User:Nathan_Reynolds>
>> | Architect | 602.333.9091
>> Oracle PSR Engineering <http://psr.us.oracle.com/> | Server Technology
>> On 4/16/2013 10:27 PM, Zhong Yu wrote:
>>> On Tue, Apr 16, 2013 at 8:51 PM, thurstonn<thurston at nomagicsoftware.com>  wrote:
>>>> Vitaly Davidovich wrote
>>>>> The code works as-is.
>>>> Absolutely.  volatile is not needed for correctness
>>>>
>>>> Vitaly Davidovich wrote
>>>>> Why?
>>>> Well, for performance reasons given the 'undefined/indefinite' visibility of
>>>> #hash to other threads.
>>>> At least according to the JMM (which has nothing to say about CPU cache
>>>> coherency), it is *possible* that each distinct thread that invoked
>>>> #hashCode() *could* result in a recalculation of the hash.
>>> In practice though, application threads contain very frequent
>>> synchronization actions, or other operations that force VM to
>>> flush/reload. So it won't take very long for any non-volatile write in
>>> one thread to become visible to other threads.
>>>> Imagine a long-lived Map<String, ?>; and many threads accessing the map's
>>>> keyset and for some unknown reason invoking #hashCode() on each key.
>>>> If #hash was declared volatile, although there is no guarantee that #hash
>>>> would only be calculated once, it is guaranteed that once a write to main
>>>> memory was completed, every *subsequent* (here meaning after the write to
>>> In JMM though, we cannot even express this guarantee. Say we have
>>> threads T1...Tn, each thread Ti burns `i` seconds CPU time first, then
>>> volatile-reads #hash, and if it's 0, calculates and volatile-writes
>>> #hash which takes 100 ns. We can find no guarantee from JMM that
>>> there's only one write; it's legal that every thread sees 0 from the
>>> volatile read.
>>>
>>> Zhong Yu
>>>
>>>> main memory) read no matter from which thread would see #hash != 0 and
>>>> therefore skip the calculation.
>>>>
>>>>
>>>>
>>>> Vitaly Davidovich wrote
>>>>> String is too high profile (especially
>>>>> hashing it) to do the "naive" thing.
>>>> Nothing wrong with being naive; naive can be charming.
>>>>
>>>>
>>>> Vitaly Davidovich wrote
>>>>> Also, some architectures pay a
>>>>> penalty for volatile loads and you'd incur that each time.
>>>> Fair point; the JDK authors only get one shot and they can't assume that
>>>> volatile reads are cheap
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> View this message in context:http://jsr166-concurrency.10961.n7.nabble.com/On-A-Formal-Definition-of-Data-Race-tp9408p9466.html
>>>> Sent from the JSR166 Concurrency mailing list archive at Nabble.com.
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>>
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