The doubling occurs when trasfering from 3.1 to 3.2. Its symmetrical
so transferring in the other direction halves the block count.
thanks in advance.
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* Columbia hasn't got an opinion worth expressing. | mc...@columbia.edu *
* "Carpe Cibum" | Marco Pineda *
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> The doubling occurs when trasfering from 3.1 to 3.2. Its symmetrical
> so transferring in the other direction halves the block count.
Sounds more like one machine has an alias with the "-k"
option set... That reports Kbytes vs 512-byte blocks.
(Or maybe the 3.1 version used -k as default???)
Note: the actual disk allocations are in 4k blocks though!
--
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Michael G. Phillips mg...@crl.com mg...@dbsoftware.com
(404)239-2766 "Just because it worked doesn't mean it works." -- me
My problem appears more complex, though, and du still seems to be the
culprit. Our product at work writes data files that du reports with
erroneous block counts. Copying the files and doing a block count on
the copied files yields the correct count.
My reading of the man page tells me that 'du' reports on block size
strictly on the basis of file size, and without regard to disk
blocking factors. How could it make a mistake in that case. All
other utilities tell me that the original file and the copy are the
same, and I have tried copying with other utilities, as well, all with
the same results. (For those who care I used cmp, dircmp and ls to
compare the files, and cp, cpio, tar and rcp to copy the files)
If there is anyone who has come across this problem, please post or
e-mail. thanks
Your application is writing a "sparse" file, either by using memory
mapped access or lseek. The unwritten blocks are not represented in
the original. When you copy the file, they are written as blocks of
zero, which is why the file grows. If you use backup/restore to copy
the original file, its "sparseness" is preserved.
--
Marc Auslander <ma...@watson.ibm.com> 914 784-6699 (Tieline 863 Fax x6306)
use:
du -k
on 3.2 to get the block size in k's.
I am used to working in k's and just alias it so that is what
I always get.