Booting from DOC2000 with GRUB loader

[John Sutton]

Nearly there ;-)...

On Thu, 02 May 2002, Russ Dill wrote:
> On Thu, 2002-05-02 at 00:47, John Sutton wrote:
> > Hi again
> > 
> > Many thanks for all that feedback!  Definitely a positive consensus
> > about the units for start_offset ;-)
> > 
> > Jasmine suggests that I should leave some extra after the end of grub for
> > future expansion, which seems a good idea, but raises in my mind the
> > question as to how grub (or anything else) knows where the nftl partition
> > starts?  Does nftl_format also write a pointer to the start block
> > somewhere else?  Or does the format contain some kind of signature?
> 
> just sure to erase any unused sectors.

Is a sector a block?  And what is a block?

Here's where I'm up to now:

./nftl_format /dev/mtd0 98304
$Id: nftl_format.c,v 1.17 2001/08/29 14:28:48 dwmw2 Exp $
Erase size not 8Kb - I'm confused

So I put a %d for meminfo.erasesize in the printf and I get:

./nftl_format /dev/mtd0 98304
$Id: nftl_format.c,v 1.17 2001/08/29 14:28:48 dwmw2 Exp $
Erase size not 8Kb - I'm confused 16384

So I've got a 16Kb erase size.  What now?  If I hack the source so that it
just falls through this test and proceeds to malloc(meminfo.erasesize * 5)
am I heading for big trouble or success?

Once I've gotten past this little gotcha, how do I go about using the
erase util to erase the piece between the end of the grub and the start of
the nftl partition?  I'm thinking that if I start the nftl at 98304 then I
don't need to erase anything because doc_loadbios will have already erased
the unoccupied 96k-87k=9k at the end of the last of the 6 blocks. 
Whereas, if I do:

./nftl_format /dev/mtd0 114688  ( i.e. 7 x 16k block )

then I'll need to do:

erase /dev/mtd0 6 1

or, should that be:

erase /dev/mtd0 98304 16384

Sorry to be a dummy ;-(

Thanks again!

***************************************************
John Sutton
SCL Internet
URL http://www.scl.co.uk/
Tel. +44 (0) 1239 711 888
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