[distcc] Using distcc on two ARM boards resulted in almost no acceleration...

Martin Pool mbp at canonical.com
Wed Aug 4 20:26:31 MDT 2010


On 3 August 2010 04:07, James W Mills <jameswmills at ti.com> wrote:
> Hello list!
>
> First of all, thanks for this great tool.  If I can get it to work for me, it will really help!
>
> Okay, I have two OMAP4430 Panda boards, both running a 2.6.34 kernel and Ubuntu 10.10 Maverick.  They are running distcc version 3.1, and gcc version 4.4.4, both native ports by Ubuntu for the arm.
>
> These two machines are connected to each other via ethernet on a private switched network.  They have an ~50Mb/s link.  I have configured distccd on the host machine (192.168.1.20) to listen to all interfaces (0.0.0.0), and to allow connections from "127.0.0.1 192.168.1.0/24".  The remote machine (192.168.1.10) is configured to allow "127.0.0.1 192.168.1.0/24" and to listen on 192.168.1.10.
>
> I am building a 2.6.34 kernel (make uImage).  I have built 4 times, twice locally, and twice "distributed".  Here are my results:
>
> make clean
> time make CC="distcc gcc" DISTCC_HOSTS="localhost 192.168.1.10" -j8 uImage
> real    44m44.648s
> user    30m4.039s
> sys     4m0.313s
>
> make clean
> time make CC="distcc gcc" DISTCC_HOSTS="localhost 192.168.1.10" -j4 uImage
> real    42m32.055s
> user    33m46.500s
> sys     3m55.273s
>
> make clean
> time make -j4 uImage
> real    45m19.444s
> user    57m2.945s
> sys     3m54.578s
>
> make clean
> time make -j2 uImage
> real    45m10.431s
> user    54m29.516s
> sys     3m56.813s
>
>
> As you can see, I saw no "real" speedup by using distcc.  Is there something I am doing wrong?

Hi James,

The first thing to do would be to set DISTCC_VERBOSE=1 and
DISTCC_LOG=something and look in the distcc log to see if it's
actually trying to distribute the jobs or  if not, why not.

I think the problem may be that setting DISTCC_HOSTS as an argument to
make only sets the make internal variable, not the environment
variable, so distcc won't see it.  Try putting it at the start of the
command line.

-- 
Martin


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