Messages in this thread | | | Subject | Re: [PATCH] bug in futex unqueue_me | From | Steven Rostedt <> | Date | Sun, 30 Jul 2006 19:53:21 -0400 |
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On Sun, 2006-07-30 at 08:38 +0200, Ingo Molnar wrote: > * Christian Borntraeger <borntrae@de.ibm.com> wrote: > > > From: Christian Borntraeger <borntrae@de.ibm.com> > > > > This patch adds a barrier() in futex unqueue_me to avoid aliasing of > > two pointers. > > > > On my s390x system I saw the following oops: > > > So the code becomes more or less: > > if (q->lock_ptr != 0) spin_lock(q->lock_ptr) > > instead of > > if (lock_ptr != 0) spin_lock(lock_ptr) > > > > Which caused the oops from above. > > interesting, how is this possible? We do a spin_lock(lock_ptr), and > taking a spinlock is an implicit barrier(). So gcc must not delay > evaluating lock_ptr to inside the critical section. And as far as i can > see the s390 spinlock implementation goes through an 'asm volatile' > piece of code, which is a barrier already. So how could this have > happened? I have nothing against adding a barrier(), but we should first > investigate why the spin_lock() didnt act as a barrier - there might be > other, similar bugs hiding. (we rely on spin_lock()s barrier-ness in a > fair number of places)
Ingo, this spinlock is probably still a barrier, but is it still a barrier on itself? That is, the problem here is that we have the compiler optimizing the lock_ptr temp variable that is used inside the spin_lock. So does a spin_lock protect itself, or just the stuff inside it?
Here we need a barrier to keep gcc from optimizing the use of the lock and not what the lock is protecting.
I don't know about other areas in the kernel that has a dynamic spin lock like this that needs protection.
-- Steve
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