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

[Stefan Hajnoczi <stefanha@linux.vnet.ibm.com>]

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.

Stefan