Mailinglist Archive: opensuse (4570 mails)

[Ken Schneider <suse-list@xxxxxxxxxxxxx>]

On Thu, 2005-11-03 at 08:06 -0500, Ken Schneider wrote:
> On Thu, 2005-11-03 at 13:37 +0100, Gerhard den Hollander wrote:
> > * Ken Schneider <suse-list@xxxxxxxxxxxxx> (Thu, Nov 03, 2005 at 07:33:01AM -0500)
> > 
> > > > grub> root (hd
> > > >  Possible disks are:  hd0 hd1
> > 
> > > > and the new disk is number 3, hdd. I can not install grub there, nor can I 
> > 
> > > I thought the third disk would be hdc, doesn't matter how it is attached
> > > to the controller.
> > 
> > 
> > Yes, but Unix is based on C, and there we start counting with 0
> > So 1st harddisk is hard disk number zero is hd0/hda
> > The disk number 3 / hd3 is then hdd
> > (which means it's the slave on the secondary IDE channel)
> > 
> So... do you have three disks or four? Disk numbering by the system
> (based on "c") starts at 0, but quantity (not based on "c") starts
> counting at 1. When I was learning math in school and the teacher held
> up four fingers and asked how many fingers we saw the answer was four
> not three.
> Hence your forth disk would be hdd "d" being the forth letter of the
> alphabet, and it would not be hd0/hda as there is no hd0, there is hda0
> meaning the first partition on the first drive.
> :-)
Oops. Even the partitions start counting at 1 as in hda1. So there is no
hd0 nor any hda0.
:-)

-- 
Ken Schneider
UNIX  since 1989, linux since 1994, SuSE  since 1998