Hi,
There are two cases where you need to know the location of the Solaris
slice (root or otherwise)
1. To locate the GRUB menu:
The slice and fdisk parttion are encoded into stage2 of GRUB so it know
which slice contains the GRUB menu.
This is true for both UFS and ZFS
2. To locate the "unix" module and the boot_archive. Note that the root
slice is *always* the slice that contains unix and boot_archive
This is in the GRUB menu entry. Specified by findroot() alone in the UFS
case and by findroot and bootfs in the ZFS case.
ZFS needs to extra step because findroot only specifies the pool and you
need either an explicit bootfs or the bootfs property
set in the pool.
pygrub appears to be a OSOL only utility. If you try SXCE (use
prtvtoc/fmthard to remove the root tag) it will boot just fine.
Vikram
John Levon wrote:
On Wed, Sep 16, 2009 at 04:09:19PM -0700, Vikram Hegde wrote:
Actually when newboot i.e. GRUB boot was introduced we did away with the
requirement that root slice needs to have the "root" tag and I booted
Solaris all the time without the root tag. AFAIK this is still true with
SXCE. I dont know if beadm (or Opensolaris) has reintroduced this
requirement.
pygrub requires it (see below). If I don't have the tag, how do I know
which slice contains the root?
regards
john
def get_solaris_slice(file, offset):
"""Find the root slice in a Solaris VTOC."""
buf = fsimage.get_disk_data(file, (offset + (DK_LABEL_LOC * SECTOR_SIZE)),
512)
if struct.unpack("<H", buf[508:510])[0] != DKL_MAGIC:
raise RuntimeError, "Invalid disklabel magic"
nslices = struct.unpack("<H", buf[30:32])[0]
for i in range(nslices):
sliceoff = 72 + 12 * i
slicetag = struct.unpack("<H", buf[sliceoff:sliceoff+2])[0]
slicesect = struct.unpack("<L", buf[sliceoff+4:sliceoff+8])[0]
if slicetag == V_ROOT:
return slicesect * SECTOR_SIZE
raise RuntimeError, "No root slice found"
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