linux - Why does bash execute 2 instances of my script when using a pipe inside of $() -


here bash script:

#!/bin/bash  $(find / -name "foo" | grep "bar")

here ps has output:

$ ps fx pid tty      stat   time command 2690 ?        sl     1:04 gnome-terminal 5903 pts/8    ss     0:00  \_ bash 7003 pts/8    s      0:00      \_ bash -x ./test_script.sh 7004 pts/8    s      0:00      |   \_ bash -x ./test_script.sh 7005 pts/8    s      0:00      |       \_ find / -name foo 7006 pts/8    s      0:00      |       \_ grep bar  $ ps aux  user       pid %cpu %mem    vsz   rss tty      stat start   time command 1000      7003  0.0  0.0   5172  1108 pts/8    s    16:23   0:00 bash -x ./test_script.sh 1000      7004  0.0  0.0   5172   520 pts/8    s    16:23   0:00 bash -x ./test_script.sh 1000      7005  0.7  0.0   4720  1176 pts/8    s    16:23   0:00 find / -name foo 1000      7006  0.0  0.0   4368   824 pts/8    s    16:23   0:00 grep bar

as can see there 2 instances of script being executed, can tell me bash doing here? why there 2 instances of script being executed , there better way of doing this?

thanks

when run subshell (the $(...) part) bash uses fork() system call creates copy of calling process (where subshell commands executed). script isn't being run again, rather command line inherited parent here since there no exec. in child shell bash sets pipeline, why see find , grep children.


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