I have a ./scr
script.
abc@~ $ cat scr
#!/bin/bash
ps
echo '-------'
echo "$(ps)"
abc@~ $
My goal is to find out how are the subprocesses created.
As far as I know, the $(...)
part should create a subshell, and therefore a new process. So the number of processes in the second call of ps
should be larger.
That's exactly how it is if I source the script in the current shell:
abc@~ $ . scr
PID TTY TIME CMD
1659 ttys000 0:00.17 -bash
-------
PID TTY TIME CMD
1659 ttys000 0:00.17 -bash
1785 ttys000 0:00.00 -bash
abc@~ $
However, when launching in an interpreting shell, the number of processes doesn't differ:
abc@~ $ ./scr
PID TTY TIME CMD
1659 ttys000 0:00.17 -bash
1790 ttys000 0:00.00 /bin/bash ./scr
-------
PID TTY TIME CMD
1659 ttys000 0:00.17 -bash
1790 ttys000 0:00.00 /bin/bash ./scr
abc@~ $
Why is it so?
Similarly, why does ps
give the same output as (ps)
?
abc@~ $ ps
PID TTY TIME CMD
1659 ttys000 0:00.18 -bash
abc@~ $ (ps)
PID TTY TIME CMD
1659 ttys000 0:00.18 -bash
abc@~ $
An interesting thing is that prepending the ps
command with any other command forces it to "produce" the expected new process (produces the expected process in the script at the top, in ./scr
, as well).
abc@~ $ (echo 1; ps)
1
PID TTY TIME CMD
1659 ttys000 0:00.20 -bash
1823 ttys000 0:00.00 -bash
abc@~ $
Is (ps)
being somehow "optimised" by the shell? And why is it not, when sourced?
A side note: the system is actually a macOS, I don't expect it to behave differently in that case, though.
EDIT:
As in this answer, the subshell seems to be a subject to optimisation, and therefore is not being run in a separate, newly initiated shell, because apparently it's not needed.
Why is it needed when running in current shell, then (. scr
)?