[...]
[...]
Okay, I understood that now. A const qualifier just forbids modifying
the underlying memory _through this particular pointer_, right?
In the case of slub's kfree(), which takes a const pointer, you pass it
on to slab_free() which actually _DOES_ modification of the underlying
memory when linking the object to the freelist (as far as I understood
the code).
So if I got it right and you actually modify the memory you only got a
const pointer to, you reach a level where you _have to_ break this
policy and cast to a non-const pointer, as it is currently done in
kfree(). No?
Hannes
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