Hi Nick,
> > That should be possible as you have the files, now without their 16
> > bytes per 512 bytes overhead. You need to work out how many blocks
> > to erase based on the size of each file you're trying to put back.
> > IIRC I pointed out a block is the unit of erasure and it's 16 KiB.
>
> So e3-nand.0 is 3584 KB so how many blocks go into it…? so how many
> times dose 16KB going in to 3584KB…?? I don’t know..?? I can’t divide.
> Iam a bit ok on Add not brilliant at Times but can’t do Divide.
Having stripped the excess OOB data, I have these sizes for each file in
bytes.
3670016 e3-nand-backup.0
262144 e3-nand-backup.1
262144 e3-nand-backup.2
262144 e3-nand-backup.3
28311552 e3-nand-backup.4
786432 e3-nand-backup.5
You're right, 3,670,016 B is 3,584 KiB because
3,670,016 / 1,024 = 3,584
so 1,024 goes into 3,670,016 exactly 3,584 times.
But the size of the erase block is 16 KiB which is
16 * 1,024 = 16,384
so we want to know how many times 16,384 goes into 3,670,016.
3,670,016 / 16,384 = 224
224 is what you worked out in your other email.
Taking those six sizes above and repeating the same sum gives
3,670,016 / 16,384 = 224
262,144 / 16,384 = 16
262,144 / 16,384 = 16
262,144 / 16,384 = 16
28,311,552 / 16,384 = 1,728
786,432 / 16,384 = 48
Now I haven't been carefully following along how to extract
e3-nand-backup.* in the first place, and how to write them back.
I'm aware there's a script involved and a command where one of the
parameters is the number of 16 KiB blocks to erase. Hopefully, plugging
in those numbers above for each of the matching areas will work, but
it's your responsibility whether to go ahead as you're the one actually
doing all this; I've just been sitting at a keyboard. :-)
--
Cheers, Ralph.
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