linux-kernel - Re: [patch 1/4] Add routine for generating an ID for kernel pointer

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Date:	Wed, 28 Dec 2011 13:47:37 +0400
From:	Pavel Emelyanov <xemul@...allels.com>
To:	Andrew Morton <akpm@...ux-foundation.org>
CC:	Cyrill Gorcunov <gorcunov@...nvz.org>,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
	Glauber Costa <glommer@...allels.com>,
	Andi Kleen <andi@...stfloor.org>, Tejun Heo <tj@...nel.org>,
	Matt Helsley <matthltc@...ibm.com>,
	Pekka Enberg <penberg@...nel.org>,
	Eric Dumazet <eric.dumazet@...il.com>,
	Vasiliy Kulikov <segoon@...nwall.com>,
	Alexey Dobriyan <adobriyan@...il.com>
Subject: Re: [patch 1/4] Add routine for generating an ID for kernel pointer

> A thought: if all we're trying to do here is to check for the sameness
> of objects, can we push the comparison into the kernel so we don't have
> this exporting-sensitive-info problem at all?  Just return a boolean to
> userspace?
> 
> Something like
> 
> int sys_pid_fields_equal(pid_t pid1, pid_t pid2, enum pid_field field_id);
> 
> ?
> 
> For /proc/pid/fdinfo/* userspace can open /proc/pid1/fdinfo/0 and
> /proc/pid2/fdinfo/0 and call sys_are_these_files_the_same(fd1, fd2, ...).
> 
> Perhaps sys_pid_fields_equal() can use sys_are_these_files_the_same()
> as well, if we can think up a way of passing it two fds to represent
> the two pids.
> 
> Have a think about it ;)

With this the complexity of determining sharing for N files scattered across
several tasks would be N^2, since we'll have to compare each file to each file.

On the other hand having just N IDs at hands would allow us to use more efficient
algorithms resulting in e.g. N*log(N) complexity.

That said I'd really appreciate if we work out a solution with IDs.

Thanks,
Pavel
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