Hello,
>>The 38G is the blocks used, an implementation detail, and the
>>93G is the file length, a property of the file.
Just wanted to confirm as --apparent-size is showing the file length. If try
to clone the file.img , do I need to keep
atleast 93G disk space emtpy or 38G ?
Thanks again....
________________________________
From: Peter Grandi <pg_...@el5.for.sabi.co.uk>
To: RH EL5 <rhelv5-list@redhat.com>
Sent: Wednesday, January 4, 2012 11:05 AM
Subject: Re: [rhelv5-list] Regarding sparse file
> Hi, If the file contains the sparse bits then can anybody
> guide me , how to remove those ? For example when I am
> checking file.img [ as below] , it's size is 38G. Basically
> it's size is 93G because of the sparse.
You are using "size" in two completely different meanings.
> [root@server]# du -sh file.img
> 38G file.img
> [root@server]# du -sh --apparent-size file.img
> 93G file.img
The 38G is the blocks used, an implementation detail, and the
93G is the file length, a property of the file.
The space occupied is usually greater than the length of the
file for several reasons, and can be smaller if it is sparse
enough (or if it is transparently compressed).
> I need the 'du -sh' and 'du-sh --apprent-size ' should be
> same and ie, 38G. Help is really appreciate.
What you seem to want is absolutely impossible. If a file is 93G
long, it is 93G long. It may only require 38G of space occupied
if it is sparse, but you cannot truncate its length to 38G
without losing information.
You can fully allocate it, or perhaps deallocate if there any
all-zero blocks withing it and release them, but that's it.
Perhaps you should state what you are trying to achieve, because
that may make sense.
Public service announcement: sparse VM disk images that are
sparse because incrementally allocated tend to perform terribly.
I have seen VM disk images created by some !"%^& with over a
million fragments as they grew from a few GiB to a hundred GiB
(that itself is a bad idea).
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