Messages in this thread | | | Date | Thu, 9 Dec 2010 10:26:59 +0100 | From | Oleg Nesterov <> | Subject | Re: blk-throttle: Correct the placement of smp_rmb() |
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On 12/08, Paul E. McKenney wrote: > > On Wed, Dec 08, 2010 at 11:06:40PM +0100, Oleg Nesterov wrote: > > > CPU0 does: > > > > A = 1; > > wmb(); > > B = 1; > > > > CPU1 does: > > > > B = 0; > > mb(); > > if (A) > > A = 2; > > > > My understanding is: after that we can safely assume that > > > > B == 1 || A == 2 > > > > IOW. Either CPU1 notices that A was changed, or CPU0 "wins" > > and sets B = 1 "after" CPU1. But, it is not possible that > > CPU1 clears B "after" it was set by CPU0 _and_ sees A == 0. > > > > Is it true? I think it should be true, but can't prove. > > I was afraid that a question like this might be coming... ;-)
I am proud of myself!
> The question is whether you can rely on the modification order of the > stores to B to deduce anything useful about the order in which the > accesses to A occurred. The answer currently is I believe you can > for a simple example such as the one above, but I am checking with > the hardware guys. In addition, please note that I am not sure if > all possible generalizations do what you want. For example, imagine a > 1024-CPU system in which the first 1023 CPUs do: > > A[smp_processor_id()] = 1; > wmb(); > B = smp_processor_id(); > > where the elements of A are cache-line aligned and padded. Suppose > that the remaining CPU does: > > i = random() % 1023; > B = -1; > mb(); > if (A[i]) > A[i] = 2; > > Are we guaranteed that B!=-1||A[i]==2? > > In this case, it could take all of the CPUs quite some time to come to > agreement on the order of all 1024 assignments to B. I am bugging some > hardware guys about this.
Yes, thanks a lot. Of course, my example was intentionally oversimplified, this generalization is closer to the real life.
> It has been awhile, so they forgot to run > away when they saw me coming. ;-)
Hehe. I am very glad you didn't run away when you saw another question from me ;)
> > CPU0: CPU1: > > > > A = 1; B = 1; > > mb(); mb(); > > if (B) if (A) > > printf("Yes"); printf("Yes"); > > > > should print "Yes" at least once. This looks very similar to > > the the previous example. > > From a hardware point of view, this example is very different than the > earlier one. You are not using the order of independent CPUs' stores to a > single variable here and in addition are using mb() everywhere instead of > a combination of mb() and wmb(). So, yes, this one is guaranteed to work.
OK, thanks.
> But what the heck are you guys really trying to do, anyway? ;-)
Vivek has already answered.
Basically, we have
update_object(obj) { modify_obj(obj);
wmb();
obj->was_changed = true; }
It can be called many times. Sooner or later, we will call
process_object(obj) { if (!obj->was_changed) return;
obj->was_changed = false;
mb();
do_process_object(obj); }
All we need is to ensure that eventually do_process_object(obj) will see the result of the last invocation of modify_obj().
IOW. It is fine to miss the changes in obj, but only if obj->was_changed continues to be T, in this case process_object() will be called again.
However, if process_object() clears ->was_changed, we must be sure that do_process_object() can't miss the result of the previous modify_obj().
Thanks Paul,
Oleg.
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