问题
I have a makefile which calls multiple other makefiles.
I'd like to pass the -j param along to the other makefile calls.
Something like (make -j8):
all:
make -f libpng_linux.mk -j$(J)
Where $(J) is the value 8 from -j8. I absolutely swear I've done this before but I cannot locate my example.
$(MAKEFLAGS) seems to contain --jobserver-fds=3,4 -j regardless of what -j2 or -j8
Edit: Possible Solution:
Will post this as an answer soon.
It appears one solution to not worry about it. Include -j8 when you call the main makefile. The sub calls to make should look like this:
all:
+make -f libpng_linux.mk -j$(J)
Notice the "+" in front of make. I noticed make tossing a warning when I tried parallel builds: make[1]: warning: jobserver unavailable: using -j1. Add `+' to parent make rule.
回答1:
Only certain flags go into $(MAKEFLAGS). -j isn't included because the sub-makes communicate with each other to ensure the appropriate number of jobs are occuring
Also, you should use $(MAKE) instead of make, since $(MAKE) will always evaluate to the correct executable name (which might not be make).
回答2:
"Do not do that" is not always the answer, but in this case it is, at least for GNU make.
GNU make parent process has an internal jobserver. If top-level Makefile is run with -j, subprocess makes will talk to the jobserver and read a parallelism level from it, without an explicit -j.
Ongoing coordination with parent's jobserver is much better for core utilization. For example, during the same build with -j6, parent could be running 2 jobs and the child 4 more, next moment both could be running 3 jobs each, then a parent would run 1 and the child 5.
来源：https://stackoverflow.com/questions/9147196/makefile-pass-jobs-param-to-sub-makefiles