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From: Hugo Mills <hugo@carfax.org.uk>
To: Eric Wheeler <btrfs@lists.ewheeler.net>
Cc: linux-btrfs@vger.kernel.org
Subject: Re: How can I get blockdev offsets of btrfs chunks for a file?
Date: Sat, 16 Jul 2016 00:17:57 +0000	[thread overview]
Message-ID: <20160716001757.GN3041@carfax.org.uk> (raw)
In-Reply-To: <alpine.LRH.2.11.1607151621220.20548@mail.ewheeler.net>

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On Fri, Jul 15, 2016 at 04:21:31PM -0700, Eric Wheeler wrote:
> Hello all,
> 
> We do btrfs subvolume snapshots over time for backups.  I would like to 
> traverse the files in the subvolumes and find the total unique chunk count 
> to calculate total space for a set of subvolumes.

   btrfs fi du may help here. Alternatively, qgroups should be able to
tell you for groups of subvols, if it's set up correctly. You
shouldn't need to implement this at a low level yourself...

> This sounds kind of like the beginning of what a deduplicator would do, 
> but I just want to count the blocks, so no submission for deduplication.  
> I started looking at bedup and other deduplicator code, but the answer to 
> this question wasn't obvious (to me, anyway).
> 
> Questions:
> 
> Is there an ioctl (or some other way) to get the block device offset for a 
> file (or file offset) so I can count the unique occurances?

   This is very much an X/Y question. There already exist a couple of
things that are at least close to the thing you actually want to
do. :)

   Hugo.

> What API documentation should I review?
> 
> Can you point me at the ioctl(s) that would handle this?
> 
> 
> Thank you for your help!
> 
> 

-- 
Hugo Mills             | Reintarnation: Coming back from the dead as a
hugo@... carfax.org.uk | hillbilly
http://carfax.org.uk/  |
PGP: E2AB1DE4          |

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      parent reply	other threads:[~2016-07-16  0:17 UTC|newest]

Thread overview: 4+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2016-07-15 23:21 How can I get blockdev offsets of btrfs chunks for a file? Eric Wheeler
2016-07-15 23:51 ` Tomasz Kusmierz
2016-07-16  0:13 ` Adam Borowski
2016-07-16  0:17 ` Hugo Mills [this message]

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