Re: [PATCH 4/4] writeback: Avoid iput() from flusher thread

From: Fengguang Wu
Date: Mon Mar 19 2012 - 07:22:36 EST


On Mon, Mar 19, 2012 at 11:46:59AM +0100, Jan Kara wrote:
> On Mon 19-03-12 04:55:15, Christoph Hellwig wrote:
> > On Fri, Mar 09, 2012 at 10:02:28AM +0100, Jan Kara wrote:
> > > Doing iput() from flusher thread (writeback_sb_inodes()) can create problems
> > > because iput() can do a lot of work - for example truncate the inode if it's
> > > the last iput on unlinked file. Some filesystems (e.g. ubifs) may need to
> > > allocate blocks during truncate (due to their COW nature) and in some cases
> > > they thus need to flush dirty data from truncate to reduce uncertainty in the
> > > amount of free space. This effectively creates a deadlock.
> > >
> > > We get rid of iput() in flusher thread by using the fact that I_SYNC inode
> > > flag effectively pins the inode in memory. So if we take care to either hold
> > > i_lock or have I_SYNC set, we can get away without taking inode reference
> > > in writeback_sb_inodes().
> > >
> > > As a side effect, we also fix possible use-after-free in wb_writeback() because
> > > inode_wait_for_writeback() call could try to reacquire i_lock on the inode that
> > > was already free.
> > >
> > > Signed-off-by: Jan Kara <jack@xxxxxxx>
> > > ---
> > > fs/fs-writeback.c | 38 ++++++++++++++++++++++++--------------
> > > fs/inode.c | 11 ++++++++++-
> > > include/linux/fs.h | 7 ++++---
> > > include/linux/writeback.h | 7 +------
> > > 4 files changed, 39 insertions(+), 24 deletions(-)
> > >
> > > diff --git a/fs/fs-writeback.c b/fs/fs-writeback.c
> > > index 1e8bf44..f9f9b61 100644
> > > --- a/fs/fs-writeback.c
> > > +++ b/fs/fs-writeback.c
> > > @@ -325,19 +325,21 @@ static int write_inode(struct inode *inode, struct writeback_control *wbc)
> > > }
> > >
> > > /*
> > > - * Wait for writeback on an inode to complete.
> > > + * Wait for writeback on an inode to complete. Called with i_lock held.
> > > + * Return 1 if we dropped i_lock and waited, 0 is returned otherwise.
> > > */
> > > -static void inode_wait_for_writeback(struct inode *inode)
> > > +int __must_check inode_wait_for_writeback(struct inode *inode)
> > > {
> > > DEFINE_WAIT_BIT(wq, &inode->i_state, __I_SYNC);
> > > wait_queue_head_t *wqh;
> > >
> > > wqh = bit_waitqueue(&inode->i_state, __I_SYNC);
> > > + if (inode->i_state & I_SYNC) {
> > > spin_unlock(&inode->i_lock);
> > > __wait_on_bit(wqh, &wq, inode_wait, TASK_UNINTERRUPTIBLE);
> > > + return 1;
> > > }
> > > + return 0;
> >
> > This is a horribly ugl primitive.
> >
> > I'd rather add a
> >
> > void inode_wait_for_writeback(struct inode *inode)
> > {
> > DEFINE_WAIT_BIT(wq, &inode->i_state, __I_SYNC);
> > wait_queue_head_t *wqh = bit_waitqueue(&inode->i_state, __I_SYNC);
> >
> > __wait_on_bit(wqh, &wq, inode_wait, TASK_UNINTERRUPTIBLE);
> > }
> >
> > and opencode all the locking ad I_SYNC checking logic in the callers.
> I agree the primitive is ugly. And actually it is buggy the way I wrote
> it. It should have been:
> __wait_on_bit(wqh, &wq, isync_wait, TASK_UNINTERRUPTIBLE);
>
> where isync_wait is:
>
> int isync_wait(void *word)
> {
> struct inode *inode = container_of(word, struct inode, i_state);
>
> spin_unlock(&inode->i_lock);
> schedule();
> return 1;
> }
>
> The problem is i_lock pins the inode for us in some cases. So once we
> drop i_lock, inode can go away so we cannot test the bit anymore.

Good point, it may not be valid to test &inode->i_state any more...

Given that __wait_on_bit() is

do {
prepare_to_wait(wq, &q->wait, mode);
if (test_bit(q->key.bit_nr, q->key.flags))
ret = (*action)(q->key.flags);
} while (test_bit(q->key.bit_nr, q->key.flags) && !ret);

The isync_wait() will do good for the first test_bit, however still
cannot avoid invalid access for the second test_bit.

The fix could be

- } while (test_bit(q->key.bit_nr, q->key.flags) && !ret);
+ } while (!ret && test_bit(q->key.bit_nr, q->key.flags));

> But there are just two places where we really need this. So maybe I can
> just opencode it there and for others use normal obvious variant.

OK.

Thanks,
Fengguang
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