Re: Why is the kfree() argument const?

[Johannes Weiner <hannes@saeurebad.de>]

Hi,

Linus Torvalds <torvalds@linux-foundation.org> writes:

[...]

> "const" is a pointer type issue, and is meant to make certain mis-uses 
> more visible at compile time. It has *no* other meaning, and anybody who 
> thinks it has is just setting himself up for problems.

[...]

>  - From a very obvious and very *real* caller perspective, "free()" really 
>    doesn't change the thing the pointer points to. It does something 
>    totally different: it makes the *pointer* itself invalid.
>
>    In other words, if you think that "kfree()" changed the thing you 
>    free'd, you're simply wrong. It did no such thing. The memory is 100% 
>    the same, it's just that you cannot access it any more, and if you try, 
>    you'll get somebody elses memory.
>
>    In other words, "kfree()" can be const.

[...]

> So never believe that "const" is some guarantee that the memory under the 
> pointer doesn't change.  That is *never* true. It has never been true in 
> C, since there can be arbitrary pointer aliases to that memory that aren't 
> actually const. If you think "const *p" means that the memory behind "p" 
> is immutable, you're simply wrong.

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