On Wed, 25 Mar 2009 12:57:51 -0400 Coleman Kane <cok...@freebsd.org> wrote:
> On Wed, 2009-03-25 at 19:30 +0300, Dmitry Marakasov wrote: > > * Pav Lucistnik (p...@freebsd.org) wrote: > > > > > > > This would break very fast -- it's passing -j3 to port Makefile > > > > > instead > > > > > of vendor Makefile. > > > > > > > > This has worked fine for me for countless years, except where the > > > > vendor's Makefiles were not parallel-safe. This has been my trick to get > > > > larger things (like mysql or xorg-server) to make in parallel. It *did* > > > > work. If this has changed, then it definitely warrants mention in > > > > UPDATING. > > > > > > Then it must have worked all these years by pure chance :) > > > > Be the way, could anyone clarify how this works? My idea was that > > passing -j to port Makefile does nothing, as make/gmake on vendor's > > Makefile is ran without any -j flags -> you get usual singlethreaded > > build. However, I have a broken port, which uses gmake and something > > like that: > > > > sometarget: > > (cd xxx; make) > > > > and that fails with -j (make: illegal option -- -). So is there > > some magic with recursive make calls and -j? > > > > When processing a Makefile, make's that support concurrent operation > look for targets that will execute the $(MAKE) program. Whenever a > compliant make is run (make or gmake), if it detects $(MAKE) in a rule > then it will automatically expand that rule into a child process that > has some sort of magical connection to the parent process. The > connection allows the different make processes to share the same pool of > "process count" resources amongst each other. > > I am not sure what the implementation is, but this communication > mechanism allows child "make" processes called with "$(MAKE) ..." from > inside a Makefile to globally only use N children (from -j N), and > otherwise block until more "free jobs" are available amongst their > shared job pool. > > I hope that's clear... Probably more so than the explanation given on > GNU make's manual. > > -- > Coleman Kane > Indeed, that is why (cd xxx; make) fails where gmake is running this command. The way GNU make and BSD make passes -j related options are different. They are not compatible. If I remember correctly, if GNU make calls BSD make (which is an error of writing Makefiles in most of vendor codes), the differences are significant and fails. Usually, changing "make" to "$(MAKE)" fixes the problems. Why did it build when "-j" was not supplied. That is because makefile rule was not GNU makefile specific such that BSD make could also executed it without problems. With "-j" switch presented, these two "make" commands became incompatible each other. I hope this helps if you guys haven't found this facts; I hope I had replied earlier. Regards, Hiro _______________________________________________ freebsd-ports@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-ports To unsubscribe, send any mail to "freebsd-ports-unsubscr...@freebsd.org"
