Use all the cores or CPUs when compiling

make -j 4
Force make command to create as many compile processes as specified (4 in the example), so that each one goes into one core or CPU and compilation happens in parallel. This reduces the time required to compile a program by up to a half in the case of CPUs with 2 cores, one fourth in the case of quad cores... and so on.

By: kovan
2009-08-05 22:50:57

What Others Think

Is there an easy way to know how many CPUs you have? Then the command could be: make -j $(cat /proc/cpus)
matthewbauer · 557 weeks and 1 day ago
Your compilation only experience a n-fold linear speedup (with n being the number of CPU/cores) if your code has only parallel components and no serial components (dependencies in your code). In the case of even a slight amount of serial components (i.e. 1-2%), speedup is greatly affected. This is the essence of Amdahl's Law.
DeusExMachina · 557 weeks and 1 day ago
@mattthewbauer in Linux you could do somethink like make -j $(grep -c ^processor /proc/cpuinfo). It doesn't do any bad to use a higher number than the actual number of cores thought. @DeusExMachina: true but usually the speed increase is linear or nearly linear, because AFAIK in Makefiles interdependencies only exist between targets, so all the source files of each target can be compiled in parallel.
kovan · 557 weeks and 1 day ago
From my make manpage, "If the -j option is given without an argument, make will not limit the number of jobs that can run simultaneously." That suggests this command shouldn't help at all. Am I wrong?
tremby · 554 weeks and 3 days ago
Oh, facepalm. I read it (more than once) as "If the -j option is not given". Never mind.
tremby · 554 weeks and 3 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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