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"/app (/dev/dsk/c0d1s0 ):51430396 blocks 3098364 files" What is the size of UFS in the command above; does block means the total space allocation?

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It would be nice in the future, if you would include the actual command that you executed. Though here it would appear to be df... it's not specified. The better question you ask, the better answer you will get! –  gabe. Aug 9 '10 at 16:12
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up vote 1 down vote accepted

From the df man page:

/export/home       (/dev/dsk/c0t0d0s7 ):  434364 blocks   108220 files

     where the columns represent  the  mount  point,  device  (or
     "filesystem",  according  to  df  -k), free blocks, and free
     files,   respectively.

As, chris specified, the -h flag will output the above in "(h)uman readable format." Other options that might be helpful are:

-b  Prints the total number of kilobytes free
-k  Prints the allocation in kbytes.

For more information, reference the df man page.

$ man df

In general, man command, will return information on the command you are executing, and how to interpret its output.

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The disk is arranged in units called "physical blocks". Physical block size for UFS is by default 512 bytes, so 51430396 free blocks is equal to 25715198 kilobytes free.

(Depending on the OS version, sometimes the -h and -k parameters aren't available so its handy to know the meaning of the output.)

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many thanks for your reply shaun –  Rajeev Sep 3 '10 at 9:59
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df -h will tell you what you want

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many thanks, found it useful –  Rajeev Aug 9 '10 at 16:03
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I've found that using df -k is better since I have found that df -h is not implemented everywhere on all OS levels. –  mdpc Aug 9 '10 at 17:42
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