linux-kernel - Re: [patch 1/7] slab: introduce kzfree()

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Date:	Tue, 24 Feb 2009 01:01:24 +1100
From:	Nick Piggin <nickpiggin@...oo.com.au>
To:	Hugh Dickins <hugh@...itas.com>
Cc:	Pekka Enberg <penberg@...helsinki.fi>,
	Matt Mackall <mpm@...enic.com>,
	KOSAKI Motohiro <kosaki.motohiro@...fujitsu.com>,
	David Vrabel <david.vrabel@....com>,
	Johannes Weiner <hannes@...xchg.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Chas Williams <chas@....nrl.navy.mil>,
	Evgeniy Polyakov <johnpol@....mipt.ru>, linux-mm@...ck.org,
	linux-kernel@...r.kernel.org,
	Christoph Lameter <cl@...ux-foundation.org>,
	Nick Piggin <npiggin@...e.de>
Subject: Re: [patch 1/7] slab: introduce kzfree()

On Friday 20 February 2009 07:36:48 Hugh Dickins wrote:
> On Thu, 19 Feb 2009, Pekka Enberg wrote:

> > Quite frankly, I fail to see how kzfree() is fundamentally different from
> > kfree(). I don't see kzfree() as a memset() + kfree() but rather as a
> > kfree() "and make sure no one sees my data". So the zeroing happens
> > _after_ you've invalidated the pointer with kzfree() so there's no
> > "zeroing of buffer going on".
>
> Well, that would be one way of picturing it, yes.
> Imagine the "z" as for "zap" rather than "zero",
> and the mechanism as opaque as Hannes suggests.
>
> > So the way I see it, Linus' argument for having const for kfree() applies
> > to kzfree().
> >
> > That said, if you guys think it's a merge blocker, by all means remove
> > the const. I just want few less open-coded ksize() users, that's all.
>
> I wouldn't call it a merge blocker, no; though I still
> think it makes far more sense without the "const" there.

Well, the buffer is only non-modified in the case of one of the
allocators (SLAB). All others overwrite some of the data region
with their own metadata.

I think it is OK to use const, though. Because k(z)free has the
knowledge that the data will not be touched by the caller any
longer.

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