A few months ago, I wrote a set of bash scripts utilizing tmux to create a simple IDE on an AIX 7.1 server. There is a bug in one of my scripts that will sometimes generate user processes very rapidly up to the limit set by ulimit. This happens very infrequently (about once per month), and I have already spent several unsuccessful hours tracking down this bug, so I decided that for the time being, I could simply set my soft user process limit to something lower than the hard limit (e.g. 100 instead of 1024) so that when my bug demonstrates itself again, there won't be a noticeable performance hit for other users on the server. Unfortunately, "ulimit -Su 100" does not appear to work in bash on AIX 7.1, but it does work in ksh. I have performed the following workaround:

Made ksh the default shell:

$ chsh [username] /usr/bin/ksh

Wrote the following to ~/.kshrc:

ulimit -Su 100  # works in ksh, but not in bash
/bin/bash -il   # start bash as an interactive login shell
exit            # once bash exits, exit from ksh, too

So now, every time I create a shell, ksh sets the soft user process limit and starts bash as an interactive login shell (I still want ~/.bash_profile to get sourced). Now I have to wonder, are the user process limits set in ksh still going to be enforced in bash subshells? In the top-level bash subshell, I ran the following:

$ ulimit -Sa
core file size          (blocks, -c) unlimited
data seg size           (kbytes, -d) unlimited
file size               (blocks, -f) unlimited
max memory size         (kbytes, -m) unlimited
open files                      (-n) unlimited
pipe size            (512 bytes, -p) 64
stack size              (kbytes, -s) unlimited
cpu time               (seconds, -t) unlimited
max user processes              (-u) 1024
virtual memory          (kbytes, -v) unlimited

As you can see, the user process limit is set to 1024.

Another major concern of mine is knowing whether or not the limit set in ksh will include processes created in tmux sessions in bash subshells.

Other detail: Whenever I create a new pane in tmux, I am fairly certain that ksh is invoked, ~/.kshrc is sourced, and bash is started, just like normal. I believe this is the case because the title of each newly created tmux panes is "ksh" (the default title of a tmux pane is the name of the current process in the foreground), yet I am presented with a bash prompt instead of a ksh prompt.

This has rambled on a bit too much, so I suppose I'll omit further detail unless asked for it.

Edit 1: Strange Behavior

Look what happens when I try to get the user process limit with "ulimit -Su" (no args), with and without using truss:

$ truss ulimit -Su 2>| truss.out
$ ulimit -Su

Maybe I'm using the tool wrong, but that looks plain strange. These commands were run inside tmux.

Edit 2: Additional Information

These commands were run from a regular bash prompt -- no subshells or tmux.

$ truss ksh -c "ulimit -Su 100" 2>&1 | grep limit
getrlimit64(9, 0x2FF1B988)                      = 0
setrlimit64(9, 0x2FF1B988)                      = 0
$ truss bash -c "ulimit -Su 100" 2>&1 | grep limit
appulimit(1005, 0)                              = 0x2001C000
bash: line 0: ulimit: max user processes: cannot modify limit: A system call received a parameter that is not valid.
  • The limits of ulimit are normally handled by the kernel so it seems that ulimit -Su works with different actual kernel limits in ksh vs bash. Could you check setrlimit() calls in both cases using truss? Dec 11, 2013 at 21:01
  • I took your suggestion and now have four different logs from running truss -u libc.a::setrlimit ulimit -Su 100 2>| truss.out from various shell environments (bash, ksh, bash tmux, and ksh tmux). What specific information should I post? Each file is approximately 150 lines long and I don't know what to look for.
    – eeowaa
    Dec 11, 2013 at 22:36
  • Did truss really do something? ulimit should not be an executable (it is a shell built-in) so truss ulimit should fail. What does which ulimit write? --- Instead you should run: truss ksh -c "ulimit -Su 100" and the same for bash. More reliable could be catching the syscall instead of the library call (-u). I would try truss ... 2>&1 | grep limit first. Dec 11, 2013 at 23:38
  • I'm not a fan of which for various reasons, including those discussed here, but which truss gives /usr/bin/ulimit. On the other hand, type ulimit gives ulimit is a shell builtin. Also, replacing ulimit with env ulimit in the truss command will work, but gives slightly different output. I will edit my question with the results of your additional suggested actions.
    – eeowaa
    Dec 12, 2013 at 0:23

1 Answer 1


Introduction to resource limits

Resource limits in Unix-like systems are controlled by getrlimit() and setrlimit() system calls. These limits are configured per process and are inherited when a new process is spawn (e.g. by fork()). It means that if you want to set a limit from a shell the command must be built into the shell (not executed as a child process). Really ulimit is a builtin in many shells including ksh and bash.

The observed behaviour

The "builtin" nature of ulimit explains the different behaviour in ksh and bash.

The limit for number of user processes (ulimit -u) is set by setrlimit(RLIMIT_NPROC, ...) syscall. In old versions of AIX RLIMIT_NPROC was not supported.<1> The support was added in AIX 6.1<2 section 5.4.4 Implemented changes> so the ulimit in ksh uses setrlimit64() correctly. The bash was probably compiled to be compatible with older version of AIX and is not able to control this limit.


You can use the ulimit builtin from ksh and all the child processes will inherit the configured limits. The shells and processes in general have nothing to do with enforcing and keeping the resource limits unless they explicitly call setrlimit().


In AIX there is also an alternative which should work in older versions of AIX too:

chdev -l sys0 -a maxuproc=100

See: 3, 4

  • Thanks for the explanation! Once I get a little more reputation on this site, I'll be sure to give this answer its deserved upvote :)
    – eeowaa
    Dec 12, 2013 at 19:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .