$$
gives process id of the script process when used in a script, like this:
Example 1
#!/bin/bash
# processid.sh
# print process ids
ps -o cmd,pid,ppid
echo "The value of $$ is $$"
$ ./processid.sh
CMD PID PPID
bash 15073 4657
/bin/bash ./processid.sh 15326 15073
ps -o cmd,pid,ppid 15327 15326
The value of $$ is 15326
Observe the pid given by $$
and ps
is 15326
My shell prompt is pid 15073
But in a subshell, $$
gives pid of parent shell (which is 15073)
Example 2
$ ( ps -o cmd,pid,ppid ; echo $$ )
CMD PID PPID
bash 15073 4657
bash 15340 15073
ps -o cmd,pid,ppid 15341 15340
15073
Here subshell is pid 15340
Question: Why so? Isn't the script also running in a subshell? What's the difference between the subshell in example 2 and the shell in which the script runs in example 1?
See Question&Answers more detail:
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