I have a question about the use of du. I understand that du reports the number of 512-byte blocks of disk usage used in the directory and below.

However, with only 3 small files (file a is 64 bytes; b is 68 bytes; c is 6 bytes), du seems to tell me that the OS is using 8 blocks x 512-byte to store them. Is there anything wrong with my calculations or I misunderstand anything?

When I do a 'du' with no option, it gives me:
8 .
When I do a 'du -k', it gives me:
4 .
When I do a 'du -a' , it gives me:
2 ./a
2 ./b
2 ./c
8 .

You have 4 small files: a, b, c, and .

The directory itself is a file and requires space to store the list "a, b, c". Each file is 1k in length, so 1k is probably the smallest chunk of disk that can be assigned to a file on your filesystem.

I guess my question is: from where I can find out the smallest chunk of disks that needs to be assigned to a file?

Also, let's say it's 1k, then, is there anyway I could change it (probably during installation or hardware changes), cus my files are mostly small and it is not efficient with it being set to 1k.

Thx.

I think that depends on how your file system works and what it supports. For e.g. vxfs (Veritas Journal) supports a block size specifier with mkfs (newfs) and also uses an extent based allocation.

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'True strength lies in gentleness' - Irish proverb