Re: [2/2] fs: Fix race between io_destroy() and io_submit() in AIO

[Milton Miller]

Hi.  I chatted with Paul a bit to clarify my thoughts.

On Tue, 15 Feb 2011 about 11:16:16 -0600, Jan Kara wrote:
> On Tue 15-02-11 12:59:24, Milton Miller wrote:
> > > A race can occur when io_submit() races with io_destroy():
> > > 
> > >  CPU1						CPU2
> > > io_submit()
> > >   do_io_submit()
> > >     ...
> > >     ctx = lookup_ioctx(ctx_id);
> > > 						io_destroy()
> > >     Now do_io_submit() holds the last reference to ctx.
> > >     ...
> > >     queue new AIO
> > >     put_ioctx(ctx) - frees ctx with active AIOs
> > > 
> > > We solve this issue by checking whether ctx is being destroyed
> > > in AIO submission path after adding new AIO to ctx. Then we
> > > are guaranteed that either io_destroy() waits for new AIO or
> > > we see that ctx is being destroyed and bail out.
> > > 
> > > Reviewed-by: Jeff Moyer <jmoyer@xxxxxxxxxx>
> > > Signed-off-by: Jan Kara <jack@xxxxxxx>
> > > CC: Nick Piggin <npiggin@xxxxxxxxx>
> > > 
> > > ---
> > > fs/aio.c |   15 +++++++++++++++
> > >  1 files changed, 15 insertions(+), 0 deletions(-)
> > > 
> > > diff --git a/fs/aio.c b/fs/aio.c
> > > index b4dd668..0244c04 100644
> > > --- a/fs/aio.c
> > > +++ b/fs/aio.c
> > > @@ -1642,6 +1642,21 @@ static int io_submit_one(struct kioctx *ctx, struct iocb __user *user_iocb,
> > >  		goto out_put_req;
> > >  
> > >  	spin_lock_irq(&ctx->ctx_lock);
> > > +	/*
> > > +	 * We could have raced with io_destroy() and are currently holding a
> > > +	 * reference to ctx which should be destroyed. We cannot submit IO
> > > +	 * since ctx gets freed as soon as io_submit() puts its reference.
> > > +	 * The check here is reliable since io_destroy() sets ctx->dead before
> > > +	 * waiting for outstanding IO. Thus if we don't see ctx->dead set here,
> > > +	 * io_destroy() waits for our IO to finish.
> > > +	 * The check is inside ctx->ctx_lock to avoid extra memory barrier
> > > +	 * in this fast path...
> > > +	 */
> > 
> > When reading this comment, and with all of the recient discussions I
> > had with Paul in the smp ipi thread (especially with resepect to third
> > party writes), I looked to see that the spinlock was paired with the
> > spinlock to set dead in io_destroy.  It is not.   It took me some time
> > to find that the paired lock is actually in wait_for_all_aios.  Also,
> > dead is also set in aio_cancel_all which is under the same spinlock.
> > 
> > Please update this lack of memory barrier comment to reflect the locking.

This locking description is wrong:

>   Hum, sorry but I don't understand. The above message wants to say that
> io_destroy() does
>   ctx->dead = 1
>   barrier (implied by a spin_unlock)

no spin_unlock only does a release barrier.

>   wait for reqs_active to get to 0

This read can move up into the spinlocked region (up to the lock acquire).

> 
> while io_submit() does
>   increment reqs_active
>   barrier (implied by a spin_lock - on a different lock but that does not
> matter as we only need the barrier semantics)

No only an acquire barrier, old writes can move into the spinlock region

>   check ctx->dead

the increment can move down past this check to the unlock here.

> 
> So if io_submit() gets past ctx->dead check, io_destroy() will certainly
> wait for our reference in reqs_active to be released.
> 
> I don't see any lock pairing needed here... But maybe I miss something.
> 
> 								Honza

spin lock and unlock are only half barriers as described in 
Documentation/memory-barriers.txt


Now, as I said, the code is ok because the active count is read and
written under ctx->ctx_lock, and aio_cancel_all sets dead under
that lock.

But the comment needs to reflect that and not just the the code is
under in some random spin_lock region instead of a memory barrier,
which is not sufficient.   Bad lock descriptions leads to making bad
code in the future, either through copying it to another context or
though future work removing the additional constraints not mentioned.

So please, comment which locks are being used here, as what
you described is not enough.


thanks,
milton
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