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 appropriate. > 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. Cheers, -- Raphaël Hertzog -- To UNSUBSCRIBE, email to debian-mentors-requ...@lists.debian.org with a subject of "unsubscribe". Trouble? Contact listmas...@lists.debian.org
