Re: free(3) const void *

["Arsen Arsenović" <arsen@gentoo.org>]

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Alejandro Colomar <alx@kernel.org> writes:

> [[PGP Signed Part:No public key for 9E8C1AFBBEFFDB32 created at 2024-01-26T16:35:04+0100 using RSA]]
> Hi Arsen,
>
> On Fri, Jan 26, 2024 at 03:24:29PM +0100, Arsen Arsenović wrote:
>> But, free() modifies the object passed to it (even if not its bit
>> representation) by freeing it.  Freeing const-passed objects would also
>> violate the constness promise, so I disagree that free should take const
>> void*.
>
> This is an interesting interpretation.  Is expiring the lifetime of an
> object a modification of the object?  Possibly.
>
> But, the standard says:
>
> 	If an attempt is made to modify an object defined with a
> 	const-qualified type through use of an lvalue with
> 	non-const-qualified type, the behavior is undefined.
>
> Even if you consider expiring the lifetime of the object a modification
> of the object, the part that says "through use of an lvalue with
> non-const-qualified type" is not fulfilled, IMO.  That would reqire
> dereferencing the pointer, to actually get the lvalue, which free(3)
> never does.

What 'free' precisely does is outside the bounds of the standard,
though.  We can assume it is permitted to so since nothing says
otherwise.

But, besides that, what I mean by 'constness promise' is that an object
must be usable following a const usage of it as if that usage never
happened.  This would certainly not be true of 'free', whether it
dereferences or not.  I am not sure if this is a formalism of the
language definition, but it is something people (and AFAIK compilers)
rely on significantly.

Have a lovely night!
--
Arsen Arsenović

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