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    SubjectRe: [PATCH] bug in futex unqueue_me
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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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