[Qemu-devel] Re: [PATCH v2 6/7] qed: Read/write support

[Avi Kivity <avi@redhat.com>]

  On 10/11/2010 12:37 PM, Stefan Hajnoczi wrote:
> On Sun, Oct 10, 2010 at 11:10:15AM +0200, Avi Kivity wrote:
> >   On 10/08/2010 05:48 PM, Stefan Hajnoczi wrote:
> >  >This patch implements the read/write state machine.  Operations are
> >  >fully asynchronous and multiple operations may be active at any time.
> >  >
> >  >Allocating writes lock tables to ensure metadata updates do not
> >  >interfere with each other.  If two allocating writes need to update the
> >  >same L2 table they will run sequentially.  If two allocating writes need
> >  >to update different L2 tables they will run in parallel.
> >  >
> >
> >  Shouldn't there be a flush between an allocating write and an L2
> >  update?  Otherwise a reuse of a cluster can move logical sectors
> >  from one place to another, causing a data disclosure.
> >
> >  Can be skipped if the new cluster is beyond the physical image size.
>
> Currently clusters are never reused and new clusters are always beyond
> physical image size.  The only exception I can think of is when the
> image file size is not a multiple of the cluster size and we round down
> to the start of cluster.
>
> In an implementation that supports TRIM or otherwise reuses clusters
> this is a cost.
>
> >  >+
> >  >+/*
> >  >+ * Table locking works as follows:
> >  >+ *
> >  >+ * Reads and non-allocating writes do not acquire locks because they do not
> >  >+ * modify tables and only see committed L2 cache entries.
> >
> >  What about a non-allocating write that follows an allocating write?
> >
> >  1 Guest writes to sector 0
> >  2 Host reads backing image (or supplies zeros), sectors 1-127
> >  3 Host writes sectors 0-127
> >  4 Guest writes sector 1
> >  5 Host writes sector 1
> >
> >  There needs to be a barrier that prevents the host and the disk from
> >  reordering operations 3 and 5, or guest operation 4 is lost.  As far
> >  as the guest is concerned no overlapping writes were issued, so it
> >  isn't required to provide any barriers.
> >
> >  (based on the comment only, haven't read the code)
>
> There is no barrier between operations 3 and 5.  However, operation 5
> only starts after operation 3 has completed because of table locking.
> It is my understanding that *independent* requests may be reordered but
> two writes to the *same* sector will not be reordered if write A
> completes before write B is issued.
>
> Imagine a test program that uses pwrite() to rewrite a counter many
> times on disk.  When the program finishes it prints the counter
> variable's last value.  This scenario is like operations 3 and 5 above.
> If we read the counter back from disk it will be the final value, not
> some intermediate value.  The writes will not be reordered.

Yes, all that is needed is a barrier in program order.  So, operation 5 
is an allocating write as long as 3 hasn't returned? (at which point it 
becomes a non-allocating write)?

-- 
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