On Sun, 13 Oct 2002, James Thompson wrote:

> I have a disklabel on a slave drive that looks similar to this:
>
> disklabel for /dev/wd1
> 8 partitions:
> #        size    offset     fstype  [fsize bsize cpg/sgs]
>   d:  19931184         0     unused      0     0         # (Cyl.    0 -
> 19772)
>   e:  19920537        63      MSDOS                      # (Cyl.    0*-
> 19762*)
> disklabel: boot block size 0
> disklabel: super block size 0
>
> Noting that there is no NetBSD part whatsoever, now I've set this label
> up as above so that I may mount the msdos partition in order to obtain
> access to files on the msdos only disk.
>
> Now was it an error on my part to have written such a label to the disk
> with disklabel -r?

You should have been prompted to "overwrite the previous contents
of the disk". From what you say below, it looks like no label was
written, for whatever reason, so no harm done.

> If I understand correctly the disklabel is normally stored inside
> partition c on i386 computers (the netbsd part of the disk) but in this
> case that part does not exist since the entire drive is devoted to
> msdos.  So my question is where has this label been written to?

I suspect it wasn't written to disk at all. If "disklabel wd1" seems
to print a label, but then sends "no disklabel" to standard error, it
means that the label you're looking at is faked (from the MBR).

> Pending the above can a disk that is entirely devoted to msdos be
> mounted under netbsd.  (This has worked but will it cause data loss on
> the msdos partition?)

In this case, yes, as the faked label is adequate. Just mount
"/dev/wd1e" as you normally would. ("Extended" partitions require
greater finesse, but fortunately, you needn't worry about that.)
Note, there are known problems with writing to NTFS partitions,
but read/write to FATxx works fine.

> I should mention that on boot I get complaints that say "wd1: has no
> disklabel" even though it does have an kernel disklabel.  Is there any
> way to get rid of this error if what I'm trying to do isn't incorrect?

No, but it's harmless.

Frederick