how to calculate correct values for resize2fs
Bodo Thiesen
bothie at gmx.de
Sat Jun 27 10:46:47 UTC 2009
Hello Jelle
1. Please don't CC to poster if you answer, we are all on this list and
get the messages anyways.
2. And please don't full-quote if not *really* necessary.
3. From your other mail:
> The problem is I have to automate this process
That's an important information you forgot to mention in the initial
post ;)
* Jelle de Jong <jelledejong at powercraft.nl> hat geschrieben:
> $ fdisk -l /dev/sda
> Disk /dev/sda: 4034 MB, 4034838528 bytes
> bc <<< "4034-16"
> resize2fs /dev/sda1 4018M
>
> # result: partition is only 983973 (4K) blocks
> # requested: 1028608 blocks
> # will not fit exiting
4034838528 is 3847 MB and not 4034 like fdisk told you. Obviously fdisk is
using that bogus SI-numbering.
4018*1024^2 is 1028608 clusters and the underlaying block device of the
file system only supports 983973 clusters (which it already provides).
So: Take the bytes value, devide it by 1024^2 to gain the real number of
MBs and subtract 18 (not 16) MB in the first call to resize2fs like this:
resize2fs -p /dev/sda1 $(echo "
scale=0;
$(
export LANG=C
fdisk -l /dev/sda | grep sda: | awk '{ print $5 }'
)/1024^2-18
" | bc)
Then shrink the partition by 16 MB then resize2fs /dev/sda1 without
argument again to fill up the unused space again.
Remember my advice from the previous mail I'v written as well.
Regards, Bodo
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