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

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Date:	Thu, 19 Feb 2009 20:36:48 +0000 (GMT)
From:	Hugh Dickins <hugh@...itas.com>
To:	Pekka Enberg <penberg@...helsinki.fi>
cc:	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 Thu, 19 Feb 2009, Pekka Enberg wrote:
> Hugh Dickins wrote:
> > 
> > But I fail to see it as a justification for kzfree(const void *):
> > if someone has "const char *string = kmalloc(size)" and then
> > wants that string zeroed before it is freed, then I think it's
> > quite right to cast out the const when calling kzfree().
> 
> 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.

Hugh
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