How do I determine the block size of an ext3 partition on Linux?

Question

How do I determine the block size of an ext3 partition on Linux?

Answer 1

# tune2fs -l /dev/sda1 | grep -i 'block size'
Block size:               1024

Replace /dev/sda1 with the partition you want to check.

Answer 2

Without root, without writing, and for any filesystem type, you can do:

stat -fc %s .

This will give block size of the filesystem mounted in current directory (or any other directory specified instead of the dot).

Answer 3

dumpe2fs -h /dev/md2

will output something with:

Block size:               4096
Fragment size:            4096

Answer 4

In the case where you don't have the right to run tune2fs on a device (e.g. in a corporate environment) you can try writing a single byte to a file on the partition in question and check the disk usage:

echo 1 > test
du -h test

Answer 5

On x86, a filesystem block is just about always 4KiB - the default size - and never larger than the size of a memory page (which is 4KiB).

Answer 6

To detect block size of required partition:

1. Detect partition name:

   $ df -h
   

   for example we have /dev/sda1

2. Detect block size for this partition:

   $ sudo blockdev --getbsz /dev/sda1
   

Answer 7

stat <<Filename>>

will also give file size in blocks

Answer 8

@narthi mentions using du -h on a tiny file too, but I'll add some more context and explanation:

How to find the cluster size of any filesystem, whether NTFS, Apple APFS, ext4, ext3, FAT, exFAT, etc.

Create a file with a single char in it, and run du -h on it to see how much disk space it takes up. This is your cluster size for your disk:

# Check cluster size by making and checking a 2-byte (1 char + null terminator I
# think) file. 
echo "1" > test.txt
# This is how many bytes this file actually *takes up* on this disk!
du -h test.txt
Check file size. This is the number of bytes in the file itself.
ls -alh test.txt | awk '{print $5}'

Example run and output, tested on Linux Ubuntu 20.04 on an ext4 filesystem. You can see here that test.txt takes up 4 KiB (4096 bytes) on the disk, since that is this disk's minimum cluster size, but its actual file size is only 2 bytes!

$ echo "1" > test.txt
$ du -h test.txt 
4.0K    test.txt
$ ls -alh test.txt | awk '{print $5}'
2

---

Another approach:

As @Mayur mentions here, you can also use stat to glean this information from our test.txt file, as shown here. The "Size" is 2 and the "IO Block" is 4096:

$ stat test.txt 
  File: test.txt
  Size: 2           Blocks: 8          IO Block: 4096   regular file
Device: fd01h/64769d    Inode: 27032142    Links: 1
Access: (0664/-rw-rw-r--)  Uid: ( 1000/ gabriel)   Gid: ( 1000/ gabriel)
Access: 2023-05-21 15:37:31.300562109 -0700
Modify: 2023-05-21 15:48:49.136721796 -0700
Change: 2023-05-21 15:48:49.136721796 -0700
 Birth: -

See also

1. If formatting a filesystem, such as exFAT, and if you have a choice on choosing the cluster size, I recommend 4 KiB, even for exFAT, which might otherwise default to something larger like 128 KiB, to keep disk usage low when you have a ton of small files. See my answer here: Is it best to reformat the hard drive to exFAT using 512kb chunk, or smaller or bigger chunks?

Answer 9

Use

sudo dumpe2fs /dev/sda1 | grep "Block size"

where /dev/sda1 is the device partition. You can get it from lsblk