Re: [PATCH] x86/refcount: Implement fast refcount_t handling

PaX Team
2017-04-24T15:16:39Z

On 24 Apr 2017 at 15:33, Peter Zijlstra wrote:

> On Mon, Apr 24, 2017 at 03:08:20PM +0200, PaX Team wrote:
> > On 24 Apr 2017 at 13:15, Peter Zijlstra wrote:
> >
> > > On Mon, Apr 24, 2017 at 01:00:18PM +0200, PaX Team wrote:
> > > > On 24 Apr 2017 at 10:32, Peter Zijlstra wrote:
> > >
> > > > > Also, you forgot nr_cpus in your bound. Afaict the worst case here is
> > > > > O(nr_tasks + 3*nr_cpus).
> > > >
> > > > what does nr_cpus have to do with winning the race?
> > >
> > > The CPUs could each run nested softirq/hardirq/nmi context poking at the
> > > refcount, irrespective of the (preempted) task context.
> >
> > that's fine but are you also assuming that the code executed in each of
> > those contexts leaks the same refcount? otherwise whatever they do to the
> > refcount is no more relevant than a non-leaking preemptible path that runs
> > to completion in a bounded amount of time (i.e., you get temporary bumps
> > and thus need to win yet another set of races to get their effects at once).
>
> For worst case analysis we have to assume it does, unless we can proof
> it doesn't. And that proof is very very hard, and would need to be
> redone every time the kernel changes.

for worst case analysis you need to show the existence of an amd64 system that
can spawn 2G tasks. then you'll have to show the feasibility of making all of
them get preempted (without a later reschedule) inside a 2 insn window.

> > that was exactly my point: all this applies to you as well. so let me ask
> > the 3rd time: what is your "argument for correctness" for a 0 refcount
> > value check? how does it prevent exploitation?
>
> What 0 count check are you talking about, the one that triggers when we
> want to increment 0 ?

are there any other 0 checks in there?

> I think I've explained that before; per reference count rules 0 means
> freed (or about to be freed when we talk RCU).

you only said the same thing, what 0 means. you (still) didn't explain how
checking for it prevents exploitation.

> The whole pattern:
>
> if (dec_and_test(&obj->ref))
> kfree(obj);
>
> expresses this etc.. Other reference counts also do this. No references
> means its getting freed.
>
> Can you agree with this?

sure, so far so good.

> If so; any attempt to increase the reference count while its (being)
> freed() is a use-after-free.

why would ever be there such an attempt? a freed object with intact memory
content is as useful for an attacker as a live one, that is, not at all.