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

[Matt Mackall]

On Thu, 2009-02-19 at 16:34 +0000, Hugh Dickins wrote:
> On Thu, 19 Feb 2009, Pekka Enberg wrote:
> > On Wed, 2009-02-18 at 10:50 +0000, David Vrabel wrote:
> > > > > Johannes Weiner wrote:
> > > > > > +void kzfree(const void *p)
> > > > > 
> > > > > Shouldn't this be void * since it writes to the memory?
> > > > 
> > > > No. kfree() writes to the memory as well to update freelists, poisoning
> > > > and such so kzfree() is not at all different from it.
> > 
> > On Thu, 2009-02-19 at 10:22 +0900, KOSAKI Motohiro wrote:
> > > I don't think so. It's debetable thing.
> > > 
> > > poisonig is transparent feature from caller.
> > > but the caller of kzfree() know to fill memory and it should know.
> > 
> > Debatable, sure, but doesn't seem like a big enough reason to make
> > kzfree() differ from kfree().
> 
> There may be more important things for us to worry about,
> but I do strongly agree with KOSAKI-san on this.
> 
> kzfree() already differs from kfree() by a "z": that "z" says please
> zero the buffer pointed to; "const" says it won't modify the buffer
> pointed to.  What sense does kzfree(const void *) make?  Why is
> keeping the declarations the same apart from the "z" desirable?
> 
> By all means refuse to add kzfree(), but please don't add it with const.
> 
> I can see that the "const" in kfree(const void *) is debatable
> [looks to see how userspace free() is defined: without a const],
> I can see that it might be nice to have some "goesaway" attribute
> for such pointers instead; but I don't see how you can argue for
> kzalloc(const void *).

This is what Linus said last time this came up:

http://lkml.org/lkml/2008/1/16/227

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