Re: Why is the kfree() argument const?

[Bernd Petrovitsch]

On Mit, 2008-01-16 at 08:48 -0800, Christoph Lameter wrote:
> On Wed, 16 Jan 2008, Johannes Weiner wrote:
> 
> > is there any reason why kfree() takes a const pointer just to degrade it
> > with the call to slab_free()/__cache_free() again?  The promise that the
> > pointee is not modified is just bogus in this case, anyway, isn't it?
> 
> The object is modified in various cases f.e. because of poisoning or the 
> need to store the free pointer. So its bogus, yes. Pekka?

Technically one should be able to pass a "const $type *" (which may
have been a "non-const $type *" before but at some point in time it
became "const $type *") to kfree().

The (formerly) constant contents as such vanishes IMHO (and it is not
really "modified").
Poisoning and free memory handling is IMHO internal stuff to the free
memory management subsystem and basically unrelated to the "life" of the
pointered contents before it's death with kfree().

	Bernd
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