No problem. 8 _bits_ gives 2^8-1 devices, which is 255.
That is over 15 times what you need for wide SCSI.
>> lun 8 == 0 3
>> partition 4 <= 3 4
>
> What about an entry for the entire disk (eg. /dev/sda)? What about
> logical partitions? If you include the entry for the entire disk, I
> use 9 entries for my primary disk.
Yes, 4 _bits_ is 2^4 entries. That gives you 15 partitions
plus the entire disk, just like now.
>> Note that the number of bits allowed is far more than normally
>> needed and is sufficient for every system reported to the kernel
>> list. The extra 4 bits can be left reserved or used for the
>> (rather artificial IMHO) distinction between disk and CD or
>> whatever.
>
> I think you are about 10 or 12 bits short.
You don't need X _bits_ for X devices. You need log2(X),
rounded up to the nearest integer.