I have a
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
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@~ $
(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.
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 (