Get pid of current subshell

Question

I am trying to get the pid of a currently executing subshell - but $$ is only returning the parent pid:

#!/usr/bin/sh

x() {
  echo "I am a subshell x echo 1 and my pid is $$"
}

y() {
  echo "I am a subshell y echo 1 and my pid is $$"
}


echo "I am the parent shell and my pid is $$"
x &
echo "Just launched x and the pid is $! "

y &
echo "Just launched y and the pid is $! "

wait

Output

I am the parent shell and my pid is 3107
Just launched x and the pid is 3108
I am a subshell x echo 1 and my pid is 3107
Just launched y and the pid is 3109
I am a subshell y echo 1 and my pid is 3107

As you can see above, when I run $$ from the function that I've backgrounded, it does not display the PID as when I do $! from the parent shell.

Answer 1

The output is correct.

Here's from the man page of bash.

Special Parameters

The shell treats several parameters specially. These parameters may only be referenced; assignment to them is not allowed.

* Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.
@ Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed).
# Expands to the number of positional parameters in decimal.
? Expands to the exit status of the most recently executed foreground pipeline.
- Expands to the current option flags as specified upon invocation, by the set builtin command, or those set by the shell itself (such as the -i option).
$ Expands to the process ID of the shell. In a () subshell, it expands to the process ID of the current shell, not the subshell.
! Expands to the process ID of the most recently executed background (asynchronous) command.

To get the PID inside of a subshell, you may use BASHPID. This is a bash only env variable.

Your new script will look like this.

#!/bin/bash

x() {
  echo "I am a subshell x echo 1 and my pid is $BASHPID"
}

y() {
  eval echo "I am a subshell y echo 1 and my pid is $BASHPID"
}


echo "I am the parent shell and my pid is $$"
x &
echo "Just launched x and the pid is $! "

y &
echo "Just launched y and the pid is $! "

wait

Answer 2

@John1024 's answer is great, but there's a little problem in there.

when a command run like (...), the command will run in a new subprocess, so

( : ; bash -c 'echo $PPID' ) 

will return process_id of (...)， not the function's process id that call (...)

if you want the function's process_id, you can just run:

$SHELL -c 'echo $PPID'

and the process_id will be outputed to stderr

if you want get the function's process_id form a variable, you can run:

$SHELL -c 'echo $PPID' | read -s func_pid

then you can get the pid from variable ${func_pid}

note: don't run this command in (...), otherwise it'll return the process_id of (...)

Answer 3

Environment:

SUSE Linux Enterprise Server 10 SP2 (i586)

GNU bash, version 3.1.17(1)-release (i586-suse-linux) Copyright (C) 2005 Free Software Foundation, Inc.

#!/usr/bin/sh

x() {
  mypid=$(awk 'BEGIN {print PROCINFO["ppid"] ; exit}')
  echo "I am a subshell x echo 1 and my pid is $mypid"
}

y() {
  mypid=$(awk 'BEGIN {print PROCINFO["ppid"] ; exit}')
  echo "I am a subshell y echo 1 and my pid is $mypid"
}


echo "I am the parent shell and my pid is $$"
x &
echo "Just launched x and the pid is $! "

y &
echo "Just launched y and the pid is $! "

wait

Result:

I am the parent shell and my pid is 27645
Just launched x and the pid is 27646
Just launched y and the pid is 27647
I am a subshell y echo 1 and my pid is 27647
I am a subshell x echo 1 and my pid is 27646

Answer 4

If you use Linux-kernel, you can use Linux-kernel's /proc/self feature to do this:

In simplest form: cd -P /proc/self && basename "${PWD}"

To keep the PWD and OLDPWD variable: PWD_BACKUP="${PWD}";OLDPWD_BACKUP="${OLDPWD}";cd -P /proc/self && basename "${PWD}";cd "${PWD_BACKUP}";OLDPWD="${OLDPWD_BACKUP}"

For example:

$ cd -P /proc/self && basename "${PWD}"
26758
$ (cd -P /proc/self && basename "${PWD}")
26959
$ (cd -P /proc/self && basename "${PWD}")
26961
$ 

Answer 5

Modern bash

If you are running bash v4 or better, the PID of the subshell is available in $BASHPID. For example:

$ echo $$ $BASHPID ; ( echo $$ $BASHPID  )
32326 32326
32326 1519

In the main shell, $BASHPID is the same as $$. In the subshell, it is updated to the subshell's PID.

Old bash (Version 3.x or Earlier)

Pre version 4, you need a workaround:

$ echo $$; ( : ; bash -c 'echo $PPID' )
11364
30279

(Hat tip: kubanczyk)

Why the colon?

Notice that, without the colon, the work-around does not work:

$ echo $$; ( bash -c 'echo $PPID' )
11364
11364

It appears that, in the above, a subshell is never created and hence the second statement returns the main shell's PID. By contrast, if we put two statements inside the parens, the subshell is created and the output is as we expect. This is true even if the other statement is a mere colon, :. In shell, the : is a no-operation: it does nothing. It does, in our case however, force the creation of the subshell which is enough to accomplish what we want.

Dash

On debian-like systems, dash is the default shell (/bin/sh). The PPID approach works for dash but with yet another twist:

$ echo $$; (  dash -c 'echo $PPID' ) 
5791
5791
$ echo $$; ( : ; dash -c 'echo $PPID' )
5791
5791
$ echo $$; (  dash -c 'echo $PPID'; : )   
5791
20961

With dash, placing the : command before the command is not sufficient but placing it after is.

POSIX

PPID is included in the POSIX specification.

Portability

mklement0 reports that the following works as is with bash, dash, and zsh but not ksh:

echo $$; (sh -c 'echo $PPID' && :)