Re: Why is the kfree() argument const?

Hi,

Linus Torvalds <torvalds@xxxxxxxxxxxxxxxxxxxx> 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
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