Re: [PATCH v2] mm: Add kvfree_sensitive() for freeing sensitive data objects

[Linus Torvalds <torvalds@linux-foundation.org>]

On Mon, Apr 6, 2020 at 12:40 PM Joe Perches <joe@perches.com> wrote:
>
> 2.1.44 changed kfree(void *) to kfree(const void *) but
> I didn't find a particular reason why.

Because "free()" should always have been const (and volatile, for that
matter, but the kernel doesn't care since we eschew volatile data
structures).

It's a bug in the C library standard.

Think of it this way: free() doesn't really change the data, it kills
the lifetime of it. You can't access it afterwards - you can neither
read it nor write it validly. That is a completely different - and
independent - operation from writing to it.

And  more importantly, it's perfectly fine to have a const data
structure (or a volatile one) that you free. The allocation may have
done something like this:

   struct mystruct {
       const struct dictionary *dictionary;
       ...
   };

and it was allocated and initialized before it was assigned to that
"dictionary" pointer. That's _good_ code.

So it wasn't const before the allocation, but it turned const
afterwards, and freeing it doesn't change that, it just kills the
lifetime entirely.

So "free()" should take a const pointer without complaining, and saying

   free(mystruct->dictionary);
   free(mystruct);

is a sensible an correct thing to do. Warning about - or requiring
that dictionary pointer to be cast to be freed - is fundamentally
wrong.

We're not bound by the fact that the C standard library got their
rules wrong, so we can fix it in the kernel.

             Linus