See: http://docs.sun.com/app/docs/doc/816-5166/6mbb1kq41?a=view
Solaris has its own scheme for dividing up the disk that is separate from partitions. Solaris can (currently) only use one partition on each disk. Solaris calls these "fdisk partitions". Within that partition (in your case partition 3) there are slices defined. So when installgrub runs with the above command, it seeks out the Solaris partition, finds the first slice (i.e. s0), and installs grub in there. The Solaris installer will put the root filesystem into s0 . As you will see in the article linked above, GRUB uses a different naming scheme for slices. Solaris/BSD slices are refered to using letters instead of numbers. The incantation is slightly easier to read: (hd0,0,a) is the first hard disk, first partition, first slice. Similarly: (hd0,0,b) is the first hard disk, first partition, second slice. So your partition 3 slice 1 would be: (hd0,2,a) So after installing grub your menu.lst should have a line that reads: root (hd0,2,a) > #installgrub /boot/grub/stage1 /boot/grub/stage2 > c0d0s0 That line should work correctly on your system, since installgrub, according to the slightly ambiguous man page, will politely ignore the presence of other partitions and dump grub into slice 0 of the Solaris partition. If you are using Windows bot manager as the primary boot loader, the easiest way to load Solaris is to use bootpart from: http://www.winimage.com/bootpart.htm Ignore the warnings about needing a FAT16 partition as the first partition - all this program does is provide a simple way of chainloading another operating system on your system using its own customised grub boot sector which is chainloaded from the Windows boot manager menu. If you are running Windows 2000 or XP, you are using Windows boot manager (which derives from Microsoft OS/2 originally) to boot your OS - it just no longer prints out its name on screen as it did on NT. Cheers Andrew. This message posted from opensolaris.org
