I've noticed on my MacBook Pro (Quad-core) that when I run make, it takes the same amount of time as make -j, and sure enough, Activity Monitor shows all four cores getting high usage. Why is this? Is there some default setting that Apple has? I mean, it would make sense for -j to be the default, but from what I've seen on the web make with no arguments should only be using one thread.
This isn't necessarily a problem, but I'd like to understand the cause nonetheless.
The -j|--jobs flag specifies/limits the number of commands that can be run simultaneously, not the number of threads to allocate to a single command. Think of this option as concurrency instead of parallelism.
For example, I can specify --jobs=2 and have both an ES6 transpiler and a SASS preprocessor running in the background, in the same terminal window, watching for any file changes I may make.
Related
I have a makefile which wraps the real build in a single recursive call, in order to grab and release a license around the build, regardless of whether the build succeeds. Example:
.PHONY main_target
main_target:
#license_grab &
#sleep 2
-#$(MAKE) real_target
#license_release
This works great, until I want to run the make in parallel. On some systems it works fine, but on other systems I run into the documented problem with the -j flag when passing options to a sub-make:
If your operating system doesn’t support the above communication, then
‘-j 1’ is always put into MAKEFLAGS instead of the value you
specified. This is because if the ‘-j’ option were passed down to
sub-makes, you would get many more jobs running in parallel than you
asked for.
Since I only ever have one recursive call, I'd really like to pass the -j flag untouched to the sub-make. I don't want to hard-code the -j flag (with a number) into the makefile, because the makefile runs on multiple machines, with differing numbers of processors. I don't want to use -j with no limit, because that launches a bunch of processes right away rather than waiting for jobs to finish. I tried using the -l flag when I build, but I found that the limit doesn't apply right away, probably because limits don't apply until make can start sampling.
Is there a way to force a sub-make to use multiple jobs? Or, a way to make the -l flag actually accomplish something?
I think I could do it using a makefile modification like using -#$(MAKE) $(JOBS) real_target, and invoking make like make JOBS="-j4" main_target.
But, I'd prefer to just use standard make parameters, not adding extra dependencies on variables. I'd also prefer to modify the makefile as little as possible.
There is no way to change this behavior on systems which don't support jobserver capabilities. Something like your JOBS variable is the only way I can think of.
I'll point out a few things: first, I assume when you say other systems you mean Windows systems: that's the only commonly-used platform I'm aware of which doesn't support the jobserver. If so, note that as of GNU make version 4.0, jobserver support is implemented on Windows. So another option is to upgrade to a newer version of GNU make.
Second, the issues with the -l option were at least partly solved in GNU make version 3.81, where an extra algorithm was introduced to artificially adjust the load average based on the number of jobs make invokes. You shouldn't see this issue any longer with versions of make after that.
The problem is that for whatever reason my make does not support the job server, thus it does not pass the -j flag on in the $(MAKEFLAGS) variable. Since upgrading make to a version that does pass the flag is not an option, the solution is to pass the -j flag elsewhere. Just like the $(MAKE) variable can be abused to pass the -f flag, it can be used in the same way to pass the -j flag:
make MAKE="make -j4" main_target
This will start the main_target build with one job, but invoke make with 4 jobs on the sub-make process. Obviously, if you need a special make tool (the normal purpose of $(MAKE) then you'll need to specify it in the MAKE string as well as in the command.
I'm playing with Go to understand its features and syntax. I've done a simple producer-consumer program with concurrent go functions, and a priority buffer in the middle. A single producer produces tasks with a certain priority and send them to a buffer using a channel. A set of consumers will ask for a task when idle, received it and consume it. The intermediate buffer will store a set of tasks in a priority queue buffer, so highest priority tasks are served first. The program also prints the Garbage collector activity (how many times it was invoked and how many time it took to collect the garbage).
I'm running it on a Raspberry Pi using Go 1.1.
The software seems to work fine but I found that at SO level, htop shows that there are 4 processes running, with the same memory use, and the sum of the CPU use is over 100% (the Raspberry Pi only has one core so I suppose it has something to do with threads/processes). Also the system load is about 7% of the CPU, I suppose because of a constant context switching at OS level. The GOMAXPROCS environment variable is set to 1 or unset.
Do you know why Go is using more than one OS process?
Code can be found here: http://pastebin.com/HJUq6sab
Thank you!
EDIT:
It seems that htop shows the lightweight processes of the system. Go programs run several of these lightweight processes (they are different from the goroutines threads) so using htop shows several processes, while psor top will show just one, as it should be.
Please try to kill all of the suspect processes and try again running it only once. Also, do not use go run, at least for now - it blurs the number of running processes at minimum.
I suspect the other instances are simply leftovers from your previous development attempts (probably invoked through go run and not properly [indirectly] killed on SIGINT [hypothesis only]), especially because there's a 1 hour "timeout" at the end of 'main' (instead of a proper synchronization or select{}). A Go binary can spawn new threads, but it should never create new processes unless explicitly asked for that. Which is not the case of your code - it doesn't even import "os/exec" or "syscall" in the first place.
If my guess about the combination of go run and using a long timeout is really the culprit, then there's possibly some difference in the RP kernel wrt what the dev guys are using for testing.
In Racket's build system, we have a build step that invokes a program that can run several parallel tasks at once. Since this is invoked from make, it would be nice to respect the -j option that make was originally invoked with.
However, as far as I can tell, there's no way to get the value of the -j option from inside the Makefile, or even as an environment variable in the programs that make invokes.
Is there a way to get this value, or the command line that make was invoked with, or something similar that would have the relevant information? It would be ok to have this only work in GNU make.
In make 4.2.1 finally they got MAKEFLAGS right. That is, you can have in your Makefile a target
opts:
#echo $(MAKEFLAGS)
and making it will tell you the value of -j parameter right.
$ make -j10 opts
-j10 --jobserver-auth=3,4
(In make 4.1 it is still broken). Needless to say, instead of echo you can invoke a script doing proper parsing of MAKEFLAGS
Note: this answer concerns make version 3.82 and earlier. For a better answer as of version 4.2, see the answer by Dima Pasechnik.
You can not tell what -j option was provided to make. Information about the number of jobs is not accessible in the regular way from make or its sub-processes, according to the following quote:
The top make and all its sub-make processes use a pipe to communicate with
each other to ensure that no more than N jobs are started across all makes.
(taken from the file called NEWS in the make 3.82 source code tree)
The top make process acts as a job server, handing out tokens to the sub-make processes via the pipe. It seems to be your goal to do your own parallel processing and still honor the indicated maximum number of simultaneous jobs as provided to make. In order to achieve that, you would somehow have to insert yourself into the communication via that pipe. However, this is an unnamed pipe and as far as I can see, there is no way for your own process to join the job-server mechanism.
By the way, the "preprocessed version of the flags" that you mention contain the expression --jobserver-fds=3,4 which is used to communicate information about the endpoints of the pipe between the make processes. This exposes a little bit of what is going on under the hood...
I want to run my make with -j8 if I'm not using distcc, but -j40 if distcc is enabled.
If I don't figure out whether or not I can use distcc until deep in the execution of the makefile, is there a way to change the -j factor at that late date? Or do I have to make the decision in a wrapper script before I invoke make? (I really don't want to run make recursively, with a different -j factor in the sub-make).
There's no way to change the number of jobs on the fly. The jobserver is configured right at the beginning of make, and it's not possible to reconfigure it with a different size without restarting make.
I'm working on a system with four logical CPS (two dual-core CPUs if it matters). I'm using make to parallelize twelve trivially parallelizable tasks and doing it from cron.
The invocation looks like:
make -k -j 4 -l 3.99 -C [dir] [12 targets]
The trouble I'm running into is that sometimes one job will finish but the next one won't startup even though it shouldn't be stopped by the load average limiter. Each target takes about four hours to complete and I'm wondering if this might be part of the problem.
Edit: Sometimes a target does fail but I use the -k option to have the rest of the make still run. I haven't noticed any correlation with jobs failing and the next job not starting.
I'd drop the '-l'
If all you plan to run the the system is this build I think the -j 4 does what you want.
Based on my memory, if you have anything else running (crond?), that can push the load average over 4.
GNU make ref
Does make think one of the targets is failing? If so, it will stop the make after the running jobs finish. You can use -k to tell it to continue even if an error occurs.
#BCS
I'm 99.9% sure that the -l isn't causeing the problem because I can watch the load average on the machine and it drops down to about three and sometimes as low as one (!) without starting the next job.