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

Nathan Reynolds nathan.reynolds at oracle.com
Wed Apr 17 02:38:59 EDT 2013


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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