What does make -jn mean?

Question

I am trying to install LLVM on my CentOS machine. In the installation tutorial of LLVM, a flag -jn is specified along with make.

It says to perform make -jn and also says "Choose n such that make doesn’t run in to swap space issue."

What is the use of the -j flag and how can I choose the value of n?

Klaatu von Schlacker · Accepted Answer · 2016-01-25T08:18:21.103

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The -j make flag denotes how many threads you want to allot for compiling.

n is, in this case, a place-holder for the number of processes.

The classic rule of thumb is that it's safe to make n = the number of cores your CPU has. So if you are on a dual core machine, you might use -j2, while on an 8-core machine -j8

In practise, I have found that to be a good starting place, but you should probably feel free to experiment a bit and see what works best for you.

edited Jan 25 '16 at 08:18

answered Jan 24 '16 at 22:46

Klaatu von Schlacker

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I like cores + 1 – WorBlux Jan 24 '16 at 23:32
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Won't it be processes not threads in a strict sense? – heemayl Jan 25 '16 at 01:27
yes good point @heemayl – Klaatu von Schlacker Jan 25 '16 at 02:58
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Cores+1 is usually a better option: it lets you have one process stalled on disk I/O while still fully utilizing your CPU. – Mark Jan 25 '16 at 04:53
`-j$(nproc)` to always give it the number of cores the system has. – Martin Braun Sep 02 '23 at 13:43

score 2 · Answer 2 · answered Jan 24 '16 at 23:01

Whether it's safe to use n = number of cores also depends on whether you have enough memory for all the parallel compile/link jobs. It could also cause issues for disk I/O. If this is a make job you only expect to need to run once it's probably better to choose a lower n and just let it take its time.

score 2 · Answer 3 · answered Jan 25 '16 at 00:58

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The -j flag tells make(1) how many processes to run in parallel. Best value depends on the tasks run, dependencies, ... a rule of thumb is the number of processors. If you give just -j (no n), make starts as many processes in parallel as possible.

answered Jan 25 '16 at 00:58

vonbrand

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score 1 · Answer 4 · answered Mar 04 '17 at 21:30

Just as a concrete example of how the -j flag affects a build operation and a reason to be careful with it:

I forgot I had MAKEFLAGS=-j5 and tried to build LLVM; once it reached 97% percent it spawned 5 ld processes at the same time, each consuming 1.5GB+ RAM. The first one crashed within 20 minutes, and the rest kept consuming more and more memory until my computer was unusable and I finally decided to Ctrl+C it and try again with a smaller -j. With -j3 my computer is still struggling with three ld processes but at least I can still type this with only a few hiccups.

Thank you for sharing this experience. It's helpful – Rick Jul 21 '20 at 09:00 — Rick, Jul 21 '20 at 09:00

What does make -jn mean?

4 Answers4