Re: [PATCH] tty: tty_mutex: fix lockdep warning intty_lock_pair(v1)

[Peter Zijlstra]

On Fri, 2012-05-25 at 15:28 +0200, Peter Zijlstra wrote:
> On Wed, 2012-05-23 at 14:01 +0800, Ming Lei wrote:
> 
> > Even though the patch is applied, there is still one related problem about
> > mixing tty_lock_pair with tty_unlock and tty_lock. If tty locks are
> > held by calling
> > tty_lock_pair, then deadlock warning between legacy_mutex/1 and legacy_mutex
> > may be triggered if tty_unlock(tty) and tty_lock(tty) are called later
> > when tty < tty2,
> > see tty_ldisc_release() in tty_release().
> 
> This just gives me a head-ache instead of explaining anything.
> 
> Having looked at the source I still don't see how it could possibly
> work,.. So the problem with tty_release() -> tty_ldisc_release() is that
> tty_ldisc_release() does an unlock/lock of tty.
> 
> However your tty_lock_pair() can still result in tty being subclass 1,
> see your else branch, nested case.
> 
> That said, how is this not a real deadlock? If you rely on tty pointer
> ordering to avoid deadlocks, you always need to lock them in the same
> order. The unlock+lock in ldisc_release violates that.
> 
> If we don't rely on the order, then why bother with the _pair()
> primitive?

A git grep reveals tty_release() is the only user of tty_lock_pair() and
while we hold tty_mutex over the tty_lock_pair() its not held over
ldisc_release().

Thus afaict we can create the following deadlock:


	cpu-A			cpu-B

lock tty_mutex
  lock tty
  lock o_tty
unlock tty_mutex

  unlock tty
			lock tty_mutex
			  lock tty
			  lock o_tty -> block on A
  lock tty -> block on B



Also, what is that plain call to schedule() doing in tty_release()?!

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