Re: [BUG] possible race between md_free_disk and md_notify_reboot

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On 20 Feb 19:55, Yu Kuai wrote:
>
> > I just take a quick look, the problem looks obviously to me, see how
> > md_seq_show() handle the iteration.
> > 
> > diff --git a/drivers/md/md.c b/drivers/md/md.c
> > index 465ca2af1e6e..7c7a58f618c1 100644
> > --- a/drivers/md/md.c
> > +++ b/drivers/md/md.c
> > @@ -9911,8 +9911,11 @@ static int md_notify_reboot(struct notifier_block
> > *this,
> >                          mddev_unlock(mddev);
> >                  }
> >                  need_delay = 1;
> > -               mddev_put(mddev);
> > -               spin_lock(&all_mddevs_lock);
> > +
> > +               spin_lock(&all_mddevs_lock)
> > +               if (atomic_dec_and_test(&mddev->active))
> > +                       __mddev_put(mddev);
> > +
> >          }
> >          spin_unlock(&all_mddevs_lock);
> 
> While cooking the patch, this is not enough, list_for_each_entry_safe()
> should be replaced with list_for_each_entry() as well.
> 
> Will send the patch soon, with:
> 
> Reported-by: Guillaume Morin <guillaume@xxxxxxxxxxx>

Thank you! I just saw the patch and we are going to test it and let you
know.

The issue with the next pointer seems to be fixed with your change.
Though I am still unclear how the 2nd potential issue I mentioned -
where the current item would be freed concurrently by mddev_free() - is
prevented. I am not finding anything in the code that seems to prevent a
concurrent call to mddev_free() for the current item in the
list_for_each_entry() loop (and therefore accessing mddev after the
kfree()).

I understand that we are getting a reference through the active atomic
in mddev_get() under the lock in md_notify_reboot() but how is that
preventing mddev_free() from freeing the mddev as soon as we release the
all_mddevs_lock in the loop?

I am not not familiar with this code so I am most likely missing
osmething but if you had the time to explain, that would be very
helpful.

TIA

Guillaume.

-- 
Guillaume Morin <guillaume@xxxxxxxxxxx>




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