$$ in a script vs $$ in a subshell

Question

$$ 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?

score 14 · Accepted Answer · answered Apr 12 '11 at 05:45

I tried and escaping (to pass the $$ to the subshell) does not work as the subshell inherits the $$ value from the parent bash. The solution to this is to use $BASHPID.

(echo $$; echo $BASHPID)

prints the PID from the parent shell and from the subshell.

Tony Delroy · Answer 2 · 2013-11-19T02:36:54.560

9

From the bash manpage:

   $      Expands  to  the  process ID of the shell.  In a () subshell, it
          expands to the process ID of the current  shell,  not  the  sub-
          shell.

edited Nov 19 '13 at 02:36

answered Apr 11 '11 at 05:43

Tony Delroy

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Now how would I get the value of $$ in the subshell, so that I get the pid of the subshell ? – Ankur Agarwal Apr 11 '11 at 22:32
See http://stackoverflow.com/questions/7994762/how-can-i-get-the-pid-of-a-subshell-in-bash3-x – Tony Delroy Nov 19 '13 at 02:35

score 6 · Answer 3 · answered Apr 11 '11 at 00:38

6

The replacement takes place in the parent shell; the subshell hasn't been started by the time the substitution takes place.

answered Apr 11 '11 at 00:38

Ignacio Vazquez-Abrams

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Can you please be little more descriptive please ? I am not able to follow what you are saying. Thanks! – Ankur Agarwal Apr 11 '11 at 04:33
@abc: Because your subshells never sees the $$. It got invoked with `( echo 1234 )` as your parent shell already substitutes the $$ with its PID. you have to escape it, so the subshell can see the $$. – flolo Apr 11 '11 at 06:45
@flolo How do I escape it. Can you please give the code ? I tried \$$, $\$. That did not do anything. – Ankur Agarwal Apr 11 '11 at 22:31
I don't think you can escape it. You can use `$BASHPID` as mentioned above, or the more portable `sh -c 'echo $PPID'`. – Idelic Apr 12 '11 at 09:09
6

Yeah, this isn't about who does the replacement; $$ *is* evaluated in the subshell, it just intentionally evaluates to the PID of the main shell as the manpage says. – Torne May 29 '12 at 15:01

score 1 · Answer 4 · answered Nov 19 '12 at 17:14

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A more portable way, linux-only, but also compatible with dash:

read -r my_pid _ < /proc/self/stat
echo $my_pid

answered Nov 19 '12 at 17:14

drizzt

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It was worring me because `$BASHPID` is only for bash! `sh -c 'echo $$;echo $$&&sleep 5&sleep 1;echo $!'` I want the second number be equals to the third, and `$BASHPID` is blank, so: `sh -c 'echo $$;read a _ – Tiago Pimenta Jan 08 '15 at 17:29

$$ in a script vs $$ in a subshell

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