On Wed, Nov 27, 2019 at 06:57:10PM -0300, Leonardo Bras wrote:
> On Wed, 2019-11-27 at 17:15 -0300, Leonardo Bras wrote:
> > > > > > So, suppose these threads, where:
> > > > > > - T1 uses a borrowed reference, and
> > > > > > - T2 is releasing the reference (close, release):
> > > > >
> > > > > Nit: T2 is releasing the *last* reference (as implied by your reference
> > > > > to close/release).
> > > >
> > > > Correct.
> > > >
> > > > > > T1 | T2
> > > > > > kvm_get_kvm() |
> > > > > > ... | kvm_put_kvm()
> > > > > > kvm_put_kvm_no_destroy() |
> > > > > >
> > > > > > The above would not trigger a use-after-free bug, but will cause a
> > > > > > memory leak. Is my above understanding right?
> > > > >
> > > > > Yes, this is correct.
> > > > >
> > > >
> > > > Then, what would not be a bug before (using kvm_put_kvm()) now is a
> > > > memory leak (using kvm_put_kvm_no_destroy()).
> > >
>
> Sorry, I missed some information on above example.
> Suppose on that example that the reorder changes take place so that
> kvm_put_kvm{,_no_destroy}() always happens after the last usage of kvm
> (in the same syscall, let's say).
That can't happen, because the ioctl() holds a reference to KVM via its
file descriptor for /dev/kvm, and ioctl() in turn prevents the fd from
being closed.
> Before T1 and T2, refcount = 1;
This is what's impossible. T1 must have an existing reference to get
into the ioctl(), and that reference cannot be dropped until the ioctl()
completes (and by completes I mean returns to userspace). Assuming no
other bugs, i.e. T2 has its own reference, then refcount >= 2.
> If T1 uses kvm_put_kvm_no_destroy():
> - T1 increases refcount (=2)
> - T2 decreases refcount (=1)
> - T1 decreases refcount, (=0) don't free kvm (memleak)
>
> If T1 uses kvm_put_kvm():
> - T1 increases refcount (= 2)
> - T2 decreases refcount (= 1)
> - T1 decreases refcount, (= 0) frees kvm.
>
> So using kvm_put_kvm_no_destroy() would introduce a memleak where it
> would have no bug.
>
> > > No, using kvm_put_kvm_no_destroy() changes how a bug would manifest, as
> > > you note below. Replacing kvm_put_kvm() with kvm_put_kvm_no_destroy()
> > > when the refcount is _guaranteed_ to be >1 has no impact on correctness.
>
> Yes, you are correct.
> But on the above case, kvm_put_kvm{,_no_destroy}() would be called
> with refcount == 1, and if reorder patch is applied, it would not cause
> any use-after-free error, even on kvm_put_kvm() case.
>
> Is the above correct?
No, see above.
