On Mon, Apr 6, 2020 at 12:40 PM Joe Perches <joe@xxxxxxxxxxx> 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
