Message-ID: <47905A95.4030500@cateee.net>
Date: Fri, 18 Jan 2008 08:51:49 +0100
From: Giacomo Catenazzi <cate@cateee.net>
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To: Linus Torvalds <torvalds@linux-foundation.org>
CC: David Schwartz <davids@webmaster.com>,
       Johannes Weiner <hannes@saeurebad.de>,
       Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
       clameter@sgi.com, penberg@cs.helsinki.fi
Subject: Re: Why is the kfree() argument const?
References: <MDEHLPKNGKAHNMBLJOLKOEIPJJAC.davids@webmaster.com> <alpine.LFD.1.00.0801171710160.2957@woody.linux-foundation.org>
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Linus Torvalds wrote:
> 
> On Thu, 17 Jan 2008, David Schwartz wrote:
>>> "const" has nothing to do with "logical state".  It has one meaning, and
>>> one meaning only: the compiler should complain if that particular type is
>>> used to do a write access.
>> Right, exactly.
> 
> So why do you complain? 
> 
> kfree() literally doesn't write to the object.
> 
>> You are the only one who has suggested it has anything to do with changes
>> through other pointers or in other ways. So you are arguing against only
>> yourself here.
> 
> No, I'm saying that "const" has absolutely *zero* meaning on writes to an 
> object through _other_ pointers (or direct access) to the object.

Hints: "restrict" is the C99 keyword for such requirement (or better
"const restrict")

BTW I think C use non const free as a BIG warning about not to be
to "smart" on optimization.

ciao
	cate
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