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Date: Tue, 29 Apr 1997 20:22:30 +1000 (EST)
From: Stephen McKay <syssgm@dtir.qld.gov.au>
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To: freebsd-hackers@freebsd.org
cc: syssgm@dtir.qld.gov.au
Subject: Re: disklabel -- owner?
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Terry Lambert <terry@lambert.org> wrote:

>> Yep, 'd' is dead (and not missed).  'c' is a bit of a left-over since the
>> slice code went in, but it's Traditional.  Lots of code knows this. :-)
>> Though, with a bit of effort we could get rid of its special purpose.
>
>If you can point us at this "lots of code", then we can use it
>instead of rewriting "disklabel", since the only serious use something
>like that would have is as a replacement for disklabel.  I really
>don't believe there's "lots of code" that knows about 'c'.  Or if
>there is, it's old, so it also knows about 'd', right?

I was thinking of the "lots of code" as a liability, not an asset.  I
mean that any code from way back (like BSD 4.2 days) will "know" about
partition 'c'.  Partition 'd' was introduced by 386BSD, and is so recent
we can ignore it.  To find the "lots of code" I was talking about, I
grep'd the source for RAW_PART (which is defined, several times, as 2).
Current code that depends on 'c' partition includes at least:

    lib/libdisk/disk.c
    lib/libdisk/write_disk.c
    sbin/ccdconfig/ccdconfig.c
    sys/dev/ccd/ccd.c
    sys/dev/vn/vn.c
    sys/i386/i386/autoconf.c
    sys/i386/isa/diskslice_machdep.c
    sys/i386/isa/fd.c
    sys/i386/isa/matcd/matcd.c
    sys/i386/isa/mcd.c
    sys/i386/isa/scd.c
    sys/i386/isa/wcd.c
    sys/i386/isa/wd.c
    sys/kern/subr_diskslice.c
    sys/pc98/pc98/atcompat_diskslice.c
    sys/pc98/pc98/diskslice_machdep.c
    sys/pc98/pc98/fd.c
    sys/pc98/pc98/wd.c
    sys/scsi/cd.c
    sys/scsi/od.c
    sys/scsi/sd.c
    sys/sys/disklabel.h
    usr.sbin/bad144/bad144.c
    usr.sbin/config/mkswapconf.c

I don't know how deeply it is embedded in these, which is why I just
called it "lots". :-)

If some soul trudges through all these converting partition 'c' knowledge
to slice knowledge, I expect a few of us, including Terry, will be pleased.

Stephen.