linux-kernel - RE: Why is the kfree() argument const?

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Message-ID: <alpine.LFD.1.00.0801171507570.2957@woody.linux-foundation.org>
Date:	Thu, 17 Jan 2008 15:10:13 -0800 (PST)
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	David Schwartz <davids@...master.com>
cc:	Johannes Weiner <hannes@...urebad.de>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	clameter@....com, penberg@...helsinki.fi
Subject: RE: Why is the kfree() argument const?

On Thu, 17 Jan 2008, David Schwartz wrote:
> 
> No, that's not what it means. It has nothing to do with memory. It has to do
> with logical state.

Blah. That's just your own made-up explanation of what you think "const"
should mean. It has no logical background or any basis in the C language.

"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.

It says nothing at all about the "logical state of the object". It cannot, 
since a single object can - and does - have multiple pointers to it.

So your standpoint not only has no relevant background to it, it's also 
not even logically consistent.

		Linus
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