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

[Peter Zijlstra <peterz@infradead.org>]

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?