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RE: [Xen-devel] blkfront/back and sector size of 2048 with CDROM's


  • To: "Keir Fraser" <Keir.Fraser@xxxxxxxxxxxx>, "xen-devel" <xen-devel@xxxxxxxxxxxxxxxxxxx>
  • From: "James Harper" <james.harper@xxxxxxxxxxxxxxxx>
  • Date: Tue, 5 Feb 2008 19:52:32 +1100
  • Delivery-date: Tue, 05 Feb 2008 00:53:02 -0800
  • List-id: Xen developer discussion <xen-devel.lists.xensource.com>
  • Thread-index: Achni2ggR2p2caN2Szy4krhWTtOFHgARf1zQAAByYdA=
  • Thread-topic: [Xen-devel] blkfront/back and sector size of 2048 with CDROM's

> The 'sectors' referenced in blkif requests are always 512 bytes. So if
you
> read 'sector-size' of 2048 from xenstore during blkif setup, that
means
> that
> the #sector alignments and sizes you send in requests must be
multiples of
> 4.

Okay... a forinstance...

I give windows a sector size of 2048 based on what I have read from
xenstore. Windows isssues a read request for sector 1 (byte offset
2048), and a length of 1 sector.

My read buffer is aligned at 1024 bytes into the page, so in the blkif
request I set nr_segments = 1, sector_number = 4 (1 * (2048 / 512)), and
in the first (and only) segment, first_sect = 2 and last_sect = 6.

Does that sound right?

Thanks

James


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