Date: Thu, 13 Mar 2014 12:46:23 -0500 From: Dan Nelson <dnelson@allantgroup.com> To: "Christopher J. Ruwe" <cjr@cruwe.de> Cc: freebsd-questions@freebsd.org Subject: Re: cannot witness rsync delta-algorithm Message-ID: <20140313174623.GA98098@dan.emsphone.com> In-Reply-To: <20140313175526.1ea13c4f@dijkstra-old.hb22.cruwe.de> References: <20140313175526.1ea13c4f@dijkstra-old.hb22.cruwe.de>
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In the last episode (Mar 13), Christopher J. Ruwe said:
> From what I understand from rsync workings, I would expect another
> rsync from 'here' to 'there' only to copy some MBs due to possibly
> different window sizes, but significantly less than 100MB. Yet, it
> does not. Why?
>
> [cjr@dijkstra:~/rsync-exp]$ rsync -vhc --inplace here there
> sent 104.88M bytes received 35 bytes 69.92M bytes/sec
> total size is 104.86M speedup is 1.00
>
> 'here' has been copied as a whole to 'there'. The '-W' flag has not
> been set!
Yes it has :) From the manpage:
-W, --whole-file
With this option rsync's delta-transfer algorithm is not used
and the whole file is sent as-is instead. The transfer may be
faster if this option is used when the bandwidth between the
source and destination machines is higher than the bandwidth
to disk (especially when the "disk" is actually a networked
filesystem). This is the default when both the source and
destination are specified as local paths, but only if no
batch-writing option is in effect.
Since the delta algorithm has to read both files and checksum them, rsync
assumes that a plain copy will be more efficient for local files. What
happens if you add --no-whole-file to your commandline?
--
Dan Nelson
dnelson@allantgroup.com
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