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Re: DEB_BUILD_OPTIONS=parallel=n work not as expected

On Sun, 06 Dec 2009, Erik Schanze wrote:
> Hi,
> [Sorry for crossposting to debian-dpkg, but perhaps they could clarify this.
> Orig. post was: http://lists.debian.org/debian-mentors/2009/12/msg00075.html]

dpkg plays zero role in that problem, -devel would have been more

> As far as I understand, the substitution comes from "make" itself
> and it occurs in every sub-make during parallel build. 

Yes, that's the reason we use $(MAKE) and not "make" IIRC.

> But is it the intended behaviour, that the make call in "build" target will
> run with option "-j" without any number?
> Because this will end in a build run with unlimited jobs, or did I miss smth.?

I don't know but $(MAKEFLAGS) doesn't need to be given on the make command
line, AFAIK it's implicit so try removing it in your invocation.

Or maybe the 2 is implicitly respected because it corresponds to the
number of fds given in --jobserver-fds.

Raphaël Hertzog

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