On Sun, Feb 14, 2010 at 04:53:40PM -0800, H. Peter Anvin wrote:
> On 02/05/2010 12:07 PM, H. Peter Anvin wrote:
> > On 01/28/2010 02:44 PM, Karel Zak wrote:
> >> On Thu, Jan 28, 2010 at 10:10:05PM +0100, Christoph Zimmermann wrote:
> >>> to be prepared for the switch to hdd with sector sizes larger than 512
> >>> byte there must be a new default to align partitions to.
> >>
> >> The plan is to use 2048 sectors (1MiB) as a default for disks with
> >> unknown topology. I'll do this change in fdisk next week and the same
> >> is planned for parted.
> >
> > I think this should be a minimum regardless of any detected disk
> > topology, except perhaps in the case of extremely small media.
>
> Let me reiterate that.
>
> I would like to suggest the following patch. Otherwise, we are using
> the minimum alignment *for the current device*, but disk layouts have a
> nasty tendency to outlive their respective media.
I'm not sure if I good understand how the layout can outlive
respective media. dd(1) to any other disk?
In some cases we cannot ignore disk topology, for example for
desktop-class 4K disks (or some NetApp setups, etc.) you have to use
alignment_offset to calculate the physical block boundary.
Currently, we (fdisk / parted) use 1MiB offset (and grain) for
devices with unknown topology only.
Karel
> diff --git a/fdisk/fdisk.c b/fdisk/fdisk.c
> index affea09..c6242c3 100644
> --- a/fdisk/fdisk.c
> +++ b/fdisk/fdisk.c
> @@ -1069,13 +1069,16 @@ update_sector_offset(void)
> * device where the offset is quarter of of whole size
> * of the device).
> */
> - unsigned long long x;
> -
> + unsigned long long x = 0;
> +
> if (has_topology)
> x = alignment_offset ? alignment_offset : io_size;
> else
> x = grain = 2048 * 512;
>
> + if (x < 2048 * 512)
> + x = 2048 * 512;
> +
> sector_offset = x / sector_size;
>
> if (total_number_of_sectors <= sector_offset * 4)
--
Karel Zak <kzak@xxxxxxxxxx>
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