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SubjectRe: [PATCH 0/2] rcu: Transform kfree_rcu() into kvfree_rcu()
On Wed, Feb 07, 2018 at 08:57:00AM -0500, Steven Rostedt wrote:
> On Wed, 7 Feb 2018 00:31:04 -0800
> "Paul E. McKenney" <paulmck@linux.vnet.ibm.com> wrote:
>
> > I see problems. We would then have two different names for exactly the
> > same thing.
> >
> > Seems like it would be a lot easier to simply document the existing
> > kfree_rcu() behavior, especially given that it apparently already works.
> > The really doesn't seem to me to be worth a name change.
>
> Honestly, I don't believe this is an RCU sub-system decision. This is a
> memory management decision.
>
> If we have kmalloc(), vmalloc(), kfree(), vfree() and kvfree(), and we

You missed kvmalloc() ...

> want kmalloc() to be freed with kfree(), and vmalloc() to be freed with
> vfree(), and for strange reasons, we don't know how the data was
> allocated we have kvfree(). That's an mm decision not an rcu one. We
> should have kfree_rcu(), vfree_rcu() and kvfree_rcu(), and honestly,
> they should not depend on kvfree() doing the same thing for everything.
> Each should call the corresponding member that they represent. Which
> would change this patch set.
>
> Why? Too much coupling between RCU and MM. What if in the future
> something changes and kvfree() goes away or changes drastically. We
> would then have to go through all the users of RCU to change them too.
>
> To me kvfree() is a special case and should not be used by RCU as a
> generic function. That would make RCU and MM much more coupled than
> necessary.

I'd still like it to be called free_rcu() ... so let's turn it around.

What memory can you allocate and then *not* free by calling kvfree()?
kvfree() can free memory allocated by kmalloc(), vmalloc(), any slab
allocation (is that guaranteed, or just something that happens to work?)
I think it can't free per-cpu allocations, bootmem, DMA allocations, or
alloc_page/get_free_page.

Do we need to be able to free any of those objects in order to rename
kfree_rcu() to just free_rcu()?

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