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What command do I use to find the size of all the files (recursively) in a Linux or Mac OS X directory?

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4 Answers 4

up vote 14 down vote accepted

The BSD version of du used in OS X reports size with 512-byte blocks -- the sizes are essentially rounded up to the next 512-byte value. This tells you the space on disk, which is larger than the amount of data. If you have a lot of small files, the difference can be large.

Here's an example.

This is the value with regular du. It's in 512-byte blocks:

$ du -s
248   .

The -h flag results in a more readable number, in kilobytes. As expected, it's half the number of 512-byte blocks:

$ du -hs
124K  .

Finally, you can use find and awk to give you the sum of actual bytes in the files. This is kind of slow, but it works:

$ find . -type f -exec ls -l {} \; | awk '{sum += $5} END {print sum}'

This value matches exactly the number reported by Finder's Get Info window. (There are no weird forks or xattrs in this set of files.) It's significantly smaller than the value reported by du.

Here's how it works: it gets a list of all the files, and passes them to ls -l; then awk is used to count up the bytes. The -type f flag is there so that only files (and not directories) get sent to ls. Without that flag, it'll also send directory names to ls, and each file will be listed twice : once as an individual file, and once as an item in the directory.

The GNU version of du can give values in actual bytes instead of blocks. It's unfortunate that the BSD version of du is not as flexible.

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Great explanation. Interestingly when using du from GNU coreutils with the -b option I get a different result than with your find pipe. For a 36 GB directory I get a 82 KB difference. –  Stefan Schmidt Jul 25 '12 at 22:15

I combined all your approuches and combined it with a human readable output the result is:

find $1 -type f -print0 | xargs -0 stat -f'%z' | awk '{b+=$1} END {print b}' | awk '{ sum=$1 ; hum[1024**3]="Gb";hum[1024**2]="Mb";hum[1024]="Kb"; for (x=1024**3; x>=1024; x/=1024){ if (sum>=x) { printf "%.2f %s\n",sum/x,hum[x];break } }}'

Link to the gist: https://gist.github.com/mlegenhausen/9365461

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du - tells the disk use not the file size.

find . -type f -print0 | xargs -0 stat -f%z | awk '{b+=$1} END {print b}'

above terminal code (im on osx 10.6) offers for me the best result and is waaay faster than "find ... -exec"

a quick benchmark

time find . -type f -print0 | xargs -0 stat -f'%z' | awk '{b+=$1} END {print b}'

real    0m0.086s
user    0m0.029s
sys 0m0.073s

time find . -type f -exec ls -l {} \; | awk '{sum += $5} END {print sum}'

real    0m18.515s
user    0m2.929s
sys 0m9.339s
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Show the size of a single file

du -h path_to_a_file

Show the size of the contents of a directory, each sub-directory, and each individual file:

du -h path_to_a_directory

Show the size of the contents of a directory:

du -sh path_to_a_directory

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du -sch if you want something more easily readable. –  Joey Aug 13 '09 at 20:09
@Johannes, I guess you should post this comment as an answer, so it can be accepted? –  Arjan Aug 15 '09 at 17:15
I find the -c option redundant (at least on Mac OS X) and du -sh supresses the subtotals. –  Daryl Spitzer Aug 17 '09 at 18:35
Aha, I didn't know one could accept one's own answer. Now at least people know the question is solved. –  Arjan Aug 18 '09 at 9:41
I'll happily accept a new answer that is an improvement on mine. –  Daryl Spitzer Aug 18 '09 at 16:10

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