Why can't I use job control in a bash script?

Solution 1:

What he meant is that job control is by default turned off in non-interactive mode (i.e. in a script.)

From the bash man page:

JOB CONTROL
       Job  control refers to the ability to selectively stop (suspend)
       the execution of processes and continue (resume) their execution at a
       later point.
       A user typically employs this facility via an interactive interface
       supplied jointly by the system’s terminal driver and bash.

and

   set [--abefhkmnptuvxBCHP] [-o option] [arg ...]
      ...
      -m      Monitor mode.  Job control is enabled.  This option is on by
              default for interactive shells on systems that support it (see
              JOB CONTROL above).  Background processes run in a separate
              process group and a line containing their exit status  is
              printed  upon  their completion.

When he said "is stupid" he meant that not only:

  1. is job control meant mostly for facilitating interactive control (whereas a script can work directly with the pid's), but also
  2. I quote his original answer, ... relies on the fact that you didn't start any other jobs previously in the script which is a bad assumption to make. Which is quite correct.

UPDATE

In answer to your comment: yes, nobody will stop you from using job control in your bash script -- there is no hard case for forcefully disabling set -m (i.e. yes, job control from the script will work if you want it to.) Remember that in the end, especially in scripting, there always are more than one way to skin a cat, but some ways are more portable, more reliable, make it simpler to handle error cases, parse the output, etc.

You particular circumstances may or may not warrant a way different from what lhunath (and other users) deem "best practices".

Solution 2:

Job control with bg and fg is useful only in interactive shells. But & in conjunction with wait is useful in scripts too.

On multiprocessor systems spawning background jobs can greatly improve the script's performance, e.g. in build scripts where you want to start at least one compiler per CPU, or process images using ImageMagick tools parallely etc.

The following example runs up to 8 parallel gcc's to compile all source files in an array:

#!bash
...
for ((i = 0, end=${#sourcefiles[@]}; i < end;)); do
    for ((cpu_num = 0; cpu_num < 8; cpu_num++, i++)); do
        if ((i < end)); then gcc ${sourcefiles[$i]} & fi
    done
    wait
done

There is nothing "stupid" about this. But you'll require the wait command, which waits for all background jobs before the script continues. The PID of the last background job is stored in the $! variable, so you may also wait ${!}. Note also the nice command.

Sometimes such code is useful in makefiles:

buildall:
    for cpp_file in *.cpp; do gcc -c $$cpp_file & done; wait

This gives much finer control than make -j.

Note that & is a line terminator like ; (write command& not command&;).

Hope this helps.