How do I determine the new size for resize2fs?

Question

I want to shrink an ext4 filesystem to make room for a new partition and came across the resize2fs program. The command looks like this:

resize2fs -p /dev/mapper/ExistingExt4 $size

How should I determine $size if I want to substract exactly 15 GiB from the current ext4 filesystem? Can I use the output of df somehow?

Answer 1

You should not use df because it shows the size as reported by the filesystem (in this case, ext4).

Use the dumpe2fs -h /dev/mapper/ExistingExt4 command to find out the real size of the partition. The -h option makes dumpe2fs show super block info without a lot other unnecessary details. From the output, you need the block count and block size.

...  
Block count:              19506168  
Reserved block count:     975308  
Free blocks:              13750966  
Free inodes:              4263842  
First block:              0  
Block size:               4096  
...

Multiplicating these values will give the partition size in bytes.

The above numbers happen to be a perfect multiple of 1024, so we can calculate the result in KiB:

$ python -c 'print 19506168.0 * 4096 / 1024'  # python2
$ python -c 'print(19506168.0 * 4096 / 1024)' # python3
78024672.0

Since you want to shrink the partition by 15 GiB (which is 15 MiB times 1 KiB):

$ python -c 'print 19506168.0 * 4096 / 1024  -  15 * 1024 * 1024'   #python2
$ python -c 'print(19506168.0 * 4096 / 1024  -  15 * 1024 * 1024)'  #python3
62296032.0

As resize2fs accepts several kinds of suffixes, one of them being K for "1024 bytes", the command for shrinking the partition to 62296032 KiB becomes:

resize2fs -p /dev/mapper/ExistingExt4 62296032K

Without unit, the number will be interpreted as a multiple of the filesystem's blocksize (4096 in this case). See man resize2fs(8)