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

Zhong Yu zhong.j.yu at gmail.com
Wed Apr 17 13:15:52 EDT 2013


On Wed, Apr 17, 2013 at 10:44 AM, Vitaly Davidovich <vitalyd at gmail.com> wrote:
> I actually expect that the optimizer *would* do this transformation on plain

I do not expect that, since println() interacts with the external
world. Consider this program:

Thread 1
[1]    shared = 1;
[2]    console.print("press enter to continue");

Thread 2
[3]    console.readLine();
[4]    r = shared;

If there is a causality from [2] to [3] in the physics world (the
human user sees the prompt which drives him to press enter), isn't it
reasonable to assume that [4] sees [1]? If we don't have that
guarantee, how can we write programs that appear to be coherent to the
outside world? Note that making [1] and [4] volatile or synchronized
won't help, as long as there is not an edge from [2] to [3].

Zhong Yu

> fields (provided it doesn't break intra-thread semantics, of course) because
> it's a perf gain.
>
> Don't know how much JIT can see through println as it ultimately calls into
> runtime and OS functions, so I'd guess it doesn't know enough or simply
> plays conservative here.  However, Nathan's point is still valid even if
> example isn't necessarily the best one.  If you had "pure" java code instead
> of an I/O call that took significant time to execute, the write would be
> delayed.  I'm not sure why that matters though for benign data races.
> Clearly if you need immediate visibility, you code for that specifically.
>
> On Apr 17, 2013 11:32 AM, "Zhong Yu" <zhong.j.yu at gmail.com> wrote:
>>
>> On Wed, Apr 17, 2013 at 1:38 AM, Nathan Reynolds
>> <nathan.reynolds at oracle.com> wrote:
>> > Couldn't JIT hoist the non-volatile writes out of the loop?
>>
>> Certainly, sorry if my statement sounds too absolute.
>>
>> > For example, the following code...
>>
>> But, is this a valid example? Can JMM really reorder around
>> System.out.println()?
>>
>> > 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 | Architect | 602.333.9091
>> > Oracle PSR Engineering | 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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