[concurrency-interest] Can a volatile read be reordered before a lazySet?

Thomas Kountis tkountis at gmail.com
Sat Dec 27 10:04:02 EST 2014


That cup of coffee must have been really good ;)
Thanks for the detailed explanation Nitsan, especially for that last piece
with the while-loop!


On Sat, Dec 27, 2014 at 8:09 AM, Nitsan Wakart <nitsanw at yahoo.com> wrote:

> Saturday morning first cup of coffee 2 cents. T&C apply.
> 1. In this case:
>   theLong.lazySet(1L);
>   Object o = theRef.get();
> There's technically nothing to stop reordering between the store and load.
> The problem is that moving the store past the load has potential
> implications to other sequences of stores to theLong/theObject which might
> race with the above. This is described in Shipilev's excellent JMM notes
> here: Java Memory Model Pragmatics (transcript)
> <http://shipilev.net/blog/2014/jmm-pragmatics/#_jmm_interpretation_roach_motel>
>
>
> [image: image]
> <http://shipilev.net/blog/2014/jmm-pragmatics/#_jmm_interpretation_roach_motel>
>
>
>
>
>
> Java Memory Model Pragmatics (transcript)
> <http://shipilev.net/blog/2014/jmm-pragmatics/#_jmm_interpretation_roach_motel>
> Happens-Before While providing a good basis to reason about programs, SO
> is not enough to construct a practical weak model. Here is why. Let us
> analyze a simple cas...
> View on shipilev.net
> <http://shipilev.net/blog/2014/jmm-pragmatics/#_jmm_interpretation_roach_motel>
> Preview by Yahoo
>
>
> IIUC this means stores and volatile loads are conservatively not reordered.
> 2. The case you quote has a further volatile read, we can break it down as:
>   long l = theLong.get() + 1; // LOADLOAD
>   theLong.lazySet(l); //STORESTORE
>   Object o = theRef.get(); // LOADLOAD
> In this case the STORE has to happen after the first load. The first load
> also must happen before the second load. The store can be reordered to
> happen after the second load though (at least in theory, but it won't due
> to 1).
> 3. The ping pong case is different, because the while loops are of
> indefinite length, potentially infinite. Delaying the store indefinitely
> would seem to break sequential consistency and so the while loop acts as an
> effective store barrier. If you replace the while loop with a for(int
> i=0;i<K;i++) loop the reordering becomes notionally feasible again.
>
>
>


-- 
Thomas Kountis
PGP: 0x069D29A3

Q: "Whats the object-oriented way to become wealthy?"
A:  Inheritance
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