How do I get the actual directory size, using UNIX/Linux standard tools?

Alternative question: How do I get du to show me the actual directory size (not disk usage)?

Since people seem to have different definitions of the term "size": My definition of "directory size" is the sum of all regular files within that directory.

I do NOT care about the size of the directory inode or whatever (blocks * block size) the files take up on the respective file system. A directory with 3 files, 1 byte each, has a directory size of 3 bytes (by my definition).

Calculating the directory size using du seems to be unreliable.
For example, mkdir foo && du -b foo reports "4096 foo", 4096 bytes instead of 0 bytes. With very large directories, the directory size reported by du -hs can be off by 100 GB (!) and more (compressed file system).

So what (tool/option) has to be used to get the actual directory size?


5 Answers 5


Here is a script displaying a human readable directory size using Unix standard tools (POSIX).

find ${1:-.} -type f -exec ls -lnq {} \+ | awk '
BEGIN {sum=0} # initialization for clarity and safety
function pp() {
  for(i=1;i<7;i++) {
    if(v<1024) break;
  printf("%.3f %sB\n", v, unit[i]);


$ ds ~        
72.891 GiB
  • And now I found another option which is missing in all suggested ls invocations here: -q. Without this option the script will break if some file name contains newline characters. Writing really reliable shell scripts is too hard… Jun 3, 2013 at 20:17
  • @SergeyVlasov The script I posted shouldn't break with such files, only merely ignoring the extra lines. The only problem case would occur should a carefully crafted file had an extra line witha fifth colon that contains a numerical value. Your suggestion would indeed avoid that situation. Thanks for the tip, script updated.
    – jlliagre
    Jun 3, 2013 at 20:47
  • Excelent answer. +1 to you sir
    – ehime
    May 29, 2014 at 14:25
  • This is one of the most reliable solutions. It works with file names that have spaces or quotes in them and it prints a human-readable size.
    – basic6
    May 2, 2015 at 18:10
  • @KIAaze Thanks for reviewing and fixing my code!
    – jlliagre
    Jun 19, 2017 at 12:46

Some versions of du support the argument --apparent-size to show apparent size instead of disk usage. So your command would be:

du -hs --apparent-size

From the man pages for du included with Ubuntu 12.04 LTS:

      print apparent sizes,  rather  than  disk  usage;  although  the
      apparent  size is usually smaller, it may be larger due to holes
      in (`sparse') files, internal  fragmentation,  indirect  blocks,
      and the like
  • 3
    does not work: report some space for empty dirs Jul 16, 2014 at 9:29
  • 1
    this worked for me.
    – connorbode
    Nov 27, 2014 at 23:37
  • 2
    It is gives significantly different sizes when you are comparing directories on different file systems. For example same folder has apparent size of 290Gb on zfs file system and 324Gb of exFat. The solutions above give same size.
    – Pixus.ru
    Sep 16, 2016 at 21:59

Assuming you have du from GNU coreutils, this command should calculate the total apparent size of arbitrary number of regular files inside a directory without any arbitrary limits on the number of files:

find . -type f -print0 | du -scb --files0-from=- | tail -n 1

Add the -l option to du if there are some hardlinked files inside, and you want to count each hardlink separately (by default du counts multiple hardlinks only once).

The most important difference with plain du -sb is that recursive du also counts sizes of directories, which are reported differently by different filesystems; to avoid this, the find command is used to pass only regular files to du. Another difference is that symlinks are ignored (if they should be counted, the find command should be adjusted).

This command will also consume more memory than plain du -sb, because using the --files0-from=FILE makes du store device and inode numbers of all processed files, as opposed to the default behavior of remembering only files with more than one hard link. (This is not an issue if the -l option is used to count hardlinks multiple times, because the only reason to store device and inode numbers is to skip hardlinked files which had been already processed.)

If you want to get a human-readable representation of the total size, just add the -h option (this works because du is invoked only once and calculates the total size itself, unlike some other suggested answers):

find . -type f -print0 | du -scbh --files0-from=- | tail -n 1

or (if you are worried that some effects of -b are then overridden by -h)

find . -type f -print0 | du -sc --apparent-size -h --files0-from=- | tail -n 1
  • 1
    Not sure what to do for FreeBSD — although -b could probably be replaced by -A -B 1, there is no equivalent for --files0-from=-, and using xargs will need some workarounds in case the file list is bigger than ARG_MAX (and some external solution for human-readable output). Jun 3, 2013 at 19:59
  • Brilliant, this way I could check that the byte-size of a Linux folder matches a Windows folder.
    – Kar.ma
    Mar 1, 2023 at 12:00
  • Thank You. After running rsync between 2 linux machines that have compression in the filesystem but are using different filesystems I kept getting slightly different sizes when I compared options of du between the filesystems. I believe the issue was the addition of the number of blocks of the folders between the 2 different filesystems. With this method I see that the number of bytes are exactly the same.
    – drescherjm
    Mar 25, 2023 at 16:28

Just an alternative, using ls:

ls -nR | grep -v '^d' | awk '{total += $5} END {print total, "Total"}'

ls -nR: -n like -l, but list numeric UIDs and GIDs and -R list subdirectories recursively.

grep -v: Invert the sense of matching, to select non-matching lines. (-v is specified by POSIX .). '^ d' will exclude the directories.

Ls command: http://linux.about.com/od/commands/l/blcmdl1_ls.htm

Man Grep: http://linux.die.net/man/1/grep


Edited as the suggestion @ Sergey Vlasov.

  • Using the -n option for ls instead of -l (show UID/GID numbers instead of names) is safer, because user and group names can contain spaces (e.g., if winbind or sssd is used to join the system to a Windows domain, you can get group names like domain users). It should also be faster due to not needing to lookup user and group names. Jun 3, 2013 at 4:49
  • Thanks, this is MUCH faster than find -exec ls!
    – gpothier
    Jun 26, 2018 at 21:44

If all you want is the size of the files, excluding the space the directories take up, you could do something like

find . -type f -print0 | xargs -0 du -scb | tail -n 1

@SergeyVlasov pointed out that this will fail if you have more files than argmax. To avoid that you could use something like:

find . -type f -exec du -sb '{}' \; | gawk '{k+=$1}END{print k}'
  • 1
    This command will silently give a wrong result if the directory contains so many files that they don't fit in the limit on execve() arguments size — in this case xargs will invoke du multiple times, and each invocation will print grand total just for its part of the complete file list, then tail will show just the total size of the last part. Jun 2, 2013 at 17:05
  • 1
    @SergeyVlasov good point, I hadn't thought of that, thanks, answer updated.
    – terdon
    Jun 2, 2013 at 17:19

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