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Message-ID: <Pine.LNX.4.44L0.0804181726490.2186-100000@iolanthe.rowland.org>
Date: Fri, 18 Apr 2008 17:45:51 -0400 (EDT)
From: Alan Stern <stern@...land.harvard.edu>
To: Peter Zijlstra <peterz@...radead.org>
cc: Kernel development list <linux-kernel@...r.kernel.org>,
Ingo Molnar <mingo@...e.hu>,
Paul E McKenney <paulmck@...ux.vnet.ibm.com>
Subject: Re: Semphore -> mutex in the device tree
On Fri, 18 Apr 2008, Peter Zijlstra wrote:
> Do I interpert this correct when I envision a call-chain like this:
>
> register_devise(A, some_parent)
> lock_device(A, NESTING_PARENT)
> D->probe()
> register_device(B, A)
> lock_device(B, NESTING_PARENT)
>
> That would work iff register_device() sets a tree-level class on B that
> is one down from A.
>
> > Or maybe I misunderstood, and you're proposing to use a node's level in
> > the tree as its lockdep nesting level.
>
> Yes, associate a class with each level like this:
>
> static struct lockdep_class_key device_tree_class[MAX_DEVICE_TREE_DEPTH];
>
> register_device(child, parent)
> {
> ...
> child->depth = parent->depth + 1;
> WARN_ON(child->depth > MAX_DEVICE_TREE_DEPTH);
> mutex_destroy(&child->lock);
> mutex_init(&child->lock);
> lockdep_set_class(&child->lock, &device_tree_class[child->depth]);
> ...
> }
Aha. I never understood that lockdep classes worked like that.
...
> So, now your sibling scenario:
>
> Lock D, E and F:
>
> mutex_lock(&D->lock);
> mutex_lock(&E->lock);
> mutex_lock_nested(&F->lock, SINGLE_DEPTH_NESTING);
>
> This will teach lockdep the following class order:
>
> device_tree_class[1]
> device_tree_class[2]
> device_tree_class[2].subclass[1]
>
> So, if at another time you do:
>
> mutex_lock(&D->lock);
> mutex_lock(&F->lock);
> mutex_lock(&E->lock, SINGLE_DEPTH_NESTING);
>
> you're still obeying that order; of course you have to somehow guarantee
> that it will never actually deadlock - otherwise you annotate a genuine
> warning away.
I'm still not sure this will end up working. In the situation I have
in mind both sibling classes are locked by the same line of code;
hence they will end up with the same nesting/subclass.
> > In that case, consider this
> > example. Suppose driver D associates a private mutex M with each of
> > its devices. Suppose D is managing device A at level 4 and device B at
> > level 5. Then we might have:
> >
> > D: Lock device B at level 5
> > D: Lock B's associated M
> >
> > (which tells lockdep that M comes after level 5), together with
> >
> > D: Lock device A at level 4
> > D: Lock A's associated M
> > D: Lock A's child at level 5
> ^ B, right?
Doesn't have to be B. It could be any device.
> > Won't this then be a lockdep violation (since M is now locked before a
> > device at level 5)?
>
> Interesting.. yes, this would make lockdep upset - basically because you
> introduce nesting of M.
>
> device_tree_class[4]
> M_class
> device_tree_class[5]
> M_class
>
> So you take M_class inside M_class.
>
> Is this a common scenario? Normally a driver would only deal with a
> single device instance at a time, so I guess that once this scenario can
> happen the driver is already aware of this, right?
I don't know if this ever occurs in the kernel. But it is a plausible
scenario.
In situations like this, where a private lock is acquired both before
and after a device lock (albeit on different levels), the most logical
solution is to assign each private lock to a class connected with the
associated device's class. Can that be made to work?
> It would need a separate annotation; if the coupling would be static
> (ps2 keyboard/mouse comes to mind) then the driver can have different
> lockdep_class_key instances.
Rather like what I just said, right?
I have wondered if it would be feasible for lockdep to support
tree-ordered locks directly. It would have to know which mutexes
belong to a tree structure, as well as the offsets to the parent
pointer and to the mutex from the beginning of the enclosing structure.
Since processes rarely lock more than a few tree members at a time,
lockdep could get away with checking a few simple rules (for downward
lock ordering):
When locking a device D, if any other device locks are
held then D's parent must already be locked. (A little
more strict than necessary, but this should be acceptable.)
When locking a device D, none of the other device locks
already held may be for descendants of D.
No doubt there are details making this less easy than I have painted
it. Still, could it reasonably be done?
Alan Stern
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