[Aboriginal] Dealing with distcc

[Rob Landley]

On Sun, Oct 7, 2012 at 1:56 PM, John Spencer
<maillist-aboriginal at barfooze.de> wrote:
> On 10/07/2012 08:45 PM, John Spencer wrote:
>>
>> i have now limited success using distcc: when i compile a bigger piece
>> of sw i get "(distcc)" output in the beginning, but in the later phase
>> it's only "gcc -foo bar.c" kinda stuff...
>
>
> well, actually it's
> distcc[203] (dcc_build_somewhere) Warning: failed to distribute, running
> locally instead
> distcc[202] (dcc_build_somewhere) Warning: failed to distribute, running
> locally instead
> distcc[204] (dcc_build_somewhere) Warning: failed to distribute, running
> locally instead

So the QEMU instance of the distcc c wrapper is trying to send it to
host, but the host distccd isn't responding, so it retries locally.

> is it possible that all aboriginal instances use the same distcc port ?
> i currently have armv4tl, armv6 and mips running...

Simultaneously, as the first thing you're trying? Try to get one
target to distribute distcc, and only add a second if that works.

In theory there's code that sets each instance up on a separate port
based on PID (in dev-environment.sh). In practice, distcc is
notoriously brittle and provides _horrible_ feedback about what it's
doing.

One of my todo items is teaching my ccwrap code (toys/ccwrap.c) to do
distcc style distribution itself. Other problems with distcc are that
if you give it a combined command line (ala "gcc hello.c -o hello
$LINKOPTS" it won't split it into "distcc gcc -c -o tempnam.o; gcc
tempnam.o -o hello $LINKOPTS" and thus the compile of something like
toybox (gcc file1.c file2.c file3.c... -o toybox") happens entirely on
the target, with no benefit from distcc at all.

I have many, many things I would like to do. I put in my two weeks
notice at work last monday because for the past year, my job has
prevented me from doing _any_ of them. I need to find a new job that
lets me put at least some time into this stuff. (Although toybox gets
priority until it hits 1.0...)

Rob

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