On 05/13, Paul E. McKenney wrote:
>
> On Tue, May 13, 2014 at 04:17:48PM +0200, Peter Zijlstra wrote:
> >
> > diff --git a/Documentation/memory-barriers.txt
> > b/Documentation/memory-barriers.txt
> > index 46412bded104..dae5158c2382 100644
> > --- a/Documentation/memory-barriers.txt
> > +++ b/Documentation/memory-barriers.txt
> > @@ -1881,9 +1881,9 @@ The whole sequence above is available in various
> > canned forms, all of which
> > event_indicated = 1;
> > wake_up_process(event_daemon);
> >
> > -A write memory barrier is implied by wake_up() and co. if and only if they
> > wake
> > -something up. The barrier occurs before the task state is cleared, and so
> > sits
> > -between the STORE to indicate the event and the STORE to set TASK_RUNNING:
> > +A full memory barrier is implied by wake_up() and co. The barrier occurs
>
> Last I checked, the memory barrier was guaranteed
I have to admit, I am confused. I simply do not understand what "memory
barrier" actually means in this discussion.
To me, wake_up/ttwu should only guarantee one thing: all the preceding
STORE's should be serialized with all the subsequent manipulations with
task->state (even with LOAD(task->state)).
> If there is a sleep-wakeup race, for example,
> between wait_event_interruptible() and wake_up(), then it looks to me
> that the following can happen:
>
> o Task A invokes wait_event_interruptible(), waiting for
> X==1.
>
> o Before Task A gets anywhere, Task B sets Y=1, does
> smp_mb(), then sets X=1.
>
> o Task B invokes wake_up(), which invokes __wake_up(), which
> acquires the wait_queue_head_t's lock and invokes
> __wake_up_common(), which sees nothing to wake up.
>
> o Task A tests the condition, finds X==1, and returns without
> locks, memory barriers, atomic instructions, or anything else
> that would guarantee ordering.
>
> o Task A then loads from Y. Because there have been no memory
> barriers, it might well see Y==0.
Sure, but I can't understand "Because there have been no memory barriers".
IOW. Suppose we add mb() into wake_up(). The same can happen anyway?
And "if a wakeup actually occurred" is not clear to me too in this context.
For example, suppose that ttwu() clears task->state but that task was not
deactivated and it is going to check the condition, do we count this as
"wakeup actually occurred" ? In this case that task still can see Y==0.
> On the other hand, if a wake_up() really does happen, then
> the fast-path out of wait_event_interruptible() is not taken,
> and __wait_event_interruptible() is called instead. This calls
> ___wait_event(), which eventually calls prepare_to_wait_event(), which
> in turn calls set_current_state(), which calls set_mb(), which does a
> full memory barrier.
Can't understand this part too... OK, and suppose that right after that
the task B from the scenario above does
Y = 1;
mb();
X = 1;
wake_up();
After that task A checks the condition, sees X==1, and returns from
wait_event() without spin_lock(wait_queue_head_t->lock) (if it also
sees list_empty_careful() == T). Then it can see Y==0 again?
> A read and a write memory barrier (-not- a full memory barrier)
> are implied by wake_up() and co. if and only if they wake
> something up.
Now this looks as if you document that, say,
X = 1;
wake_up();
Y = 1;
doesn't need wmb() before "Y = 1" if wake_up() wakes something up. Do we
really want to document this? Is it fine to rely on this guarantee?
> The write barrier occurs before the task state is
> cleared, and so sits between the STORE to indicate the event and
> the STORE to set TASK_RUNNING, and the read barrier after that:
Plus: between the STORE to indicate the event and the LOAD which checks
task->state, otherwise:
> CPU 1 CPU 2
> =============================== ===============================
> set_current_state(); STORE event_indicated
> set_mb(); wake_up();
> STORE current->state <write barrier>
> <general barrier> STORE current->state
> LOAD event_indicated <read barrier>
this code is still racy.
In short: I am totally confused and most probably misunderstood you ;)
Oleg.
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