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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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