Re: [PATCH 10/12] fs/locks: create a tree of dependent requests.

From: NeilBrown <neilb@suse.com>
To: "J. Bruce Fields" <bfields@fieldses.org>
Cc: Jeff Layton <jlayton@kernel.org>,
	Alexander Viro <viro@zeniv.linux.org.uk>,
	Martin Wilck <mwilck@suse.de>,
	linux-fsdevel@vger.kernel.org,
	Frank Filz <ffilzlnx@mindspring.com>,
	linux-kernel@vger.kernel.org
Subject: Re: [PATCH 10/12] fs/locks: create a tree of dependent requests.
Date: Fri, 09 Nov 2018 17:24:04 +1100	[thread overview]
Message-ID: <87bm6ywyyj.fsf@notabene.neil.brown.name> (raw)
In-Reply-To: <20181109030913.GA8831@fieldses.org>

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On Thu, Nov 08 2018, J. Bruce Fields wrote:

> On Fri, Nov 09, 2018 at 11:38:19AM +1100, NeilBrown wrote:
>> On Thu, Nov 08 2018, J. Bruce Fields wrote:
>> 
>> > On Mon, Nov 05, 2018 at 12:30:48PM +1100, NeilBrown wrote:
>> >> When we find an existing lock which conflicts with a request,
>> >> and the request wants to wait, we currently add the request
>> >> to a list.  When the lock is removed, the whole list is woken.
>> >> This can cause the thundering-herd problem.
>> >> To reduce the problem, we make use of the (new) fact that
>> >> a pending request can itself have a list of blocked requests.
>> >> When we find a conflict, we look through the existing blocked requests.
>> >> If any one of them blocks the new request, the new request is attached
>> >> below that request, otherwise it is added to the list of blocked
>> >> requests, which are now known to be mutually non-conflicting.
>> >> 
>> >> This way, when the lock is released, only a set of non-conflicting
>> >> locks will be woken, the rest can stay asleep.
>> >> If the lock request cannot be granted and the request needs to be
>> >> requeued, all the other requests it blocks will then be woken
>> >
>> > So, to make sure I understand: the tree of blocking locks only ever has
>> > three levels (the active lock, the locks blocking on it, and their
>> > children?)
>> 
>> Not correct.
>> Blocks is only vertical, never horizontal.  Siblings never block each
>> other.
>> So one process hold a lock on a byte, and 27 other process want a lock
>> on that byte, then there will be 28 levels in a narrow tree - it is
>> effectively a queue.
>> Branching (via siblings) only happens when a child conflict with only
>> part of the lock held by the parent.
>> So if one process locks 32K, then two other processes request locks on
>> the 2 16K halves, then 4 processes request locks on the 8K quarters, and
>> so-on, then you could end up with 32767 processes in a binary tree, with
>> half of them all waiting on different individual bytes.
>
> Maybe I should actually read the code carefully instead of just skimming
> the changelog and jumping to conclusions.
>
> I think this is correct, but I wish we had an actual written-out
> argument that it's correct, because intuition isn't a great guide for
> posix file locks.
>
> Maybe:
>
> Waiting and applied locks are all kept in trees whose properties are:
>
> 	- the root of a tree may be an applied or unapplied lock.
> 	- every other node in the tree is an unapplied lock that
> 	  conflicts with every ancestor of that node.
>
> Every such tree begins life as an unapplied singleton which obviously
> satisfies the above properties.
>
> The only ways we modify trees preserve these properties:
>
> 	1. We may add a new child, but only after first verifying that it
> 	   conflicts with all of its ancestors.
> 	2. We may remove the root of a tree, creating a new singleton
> 	   tree from the root and N new trees rooted in the immediate
> 	   children.
> 	3. If the root of a tree is not currently an applied lock, we may
> 	   apply it (if possible).
> 	4. We may upgrade the root of the tree (either extend its range,
> 	   or upgrade its entire range from read to write).
>
> When an applied lock is modified in a way that reduces or downgrades any
> part of its range, we remove all its children (2 above).
>
> For each of those child trees: if the root of the tree applies, we do so
> (3).  If it doesn't, it must conflict with some applied lock.  We remove
> all of its children (2), and add it is a new leaf to the tree rooted in
> the applied lock (1).  We then repeat the process recursively with those
> children.
>

Thanks pretty thorough - and even looks correct.
I'll re-reading some time when it isn't late, and maybe make it into a
comment in the code.
I agree, this sort of documentation can be quite helpful.

Thanks,
NeilBrown

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