I want to start process (eg. myCommand) and get its pid (to allow to kill it later).

I tried ps and filter by name, but I can not distinguish process by names

ps ux | awk '/<myCommand>/ {print $2}' 

Because processes names are not unique.

I can run process by:

myCommand &

I found that I can get this PID by:

echo $!

Is there any simpler solution?

I would be happy to execute myCommand and get its PID as a result of one line command.


8 Answers 8


What can be simpler than echo $!? As one line:

myCommand & echo $!
  • 2
    Thank you merging these commands with "&" helped me a lot.
    – rafalmag
    Commented Nov 27, 2010 at 12:09
  • 11
    in a bash script, in a loop which starts programs, $! is not accurate. Sometimes it returns pid of the script itself, sometimes of grep or awk run from the script any solution to specifically get the pid of the process just launched in this scenario? something like pid=myprogram would have been awesome
    – user217730
    Commented Apr 26, 2014 at 5:59
  • 4
    NB that this requires you to start the command using a & as the previous line, otherwise the echo basically returns blank.
    – rogerdpack
    Commented Feb 4, 2015 at 18:56
  • 3
    assigning to variable like command & echo $! freezes the execution at this step :( Commented Jul 15, 2016 at 9:41
  • This doesn't work if you don't explicitly background the process! For instance, your process may background itself after it has done something you need to wait for before continuing.
    – Michael
    Commented Apr 16, 2020 at 0:10

You can use sh -c and exec to get the command's PID even before it runs.

To start myCommand, so that its PID is printed before it begins to run, you can use:

sh -c 'echo $$; exec myCommand'

How it works:

This starts a new shell, prints the PID of that shell, and then uses the exec builtin to replace the shell with your command, ensuring it has the same PID. When your shell runs a command with the exec builtin, your shell is actually becoming that command, rather than the more common behavior of forking a new copy of itself, which has its own separate PID and which then becomes the command.

I find this to be much simpler than alternatives involving asynchronous execution (with &), job control, or searching with ps. Those approaches are fine, but unless you have a specific reason to use them--for example, perhaps the command is already running, in which case searching for its PID or using job control would make sense--I suggest considering this way first. (And I would certainly not consider writing a complex script or other program to achieve this).

This answer includes an example of this technique.

Parts of that command could occasionally be omitted, but not usually.

Even if the shell you're using is a Bourne-style and thus supports the exec builtin with these semantics, you generally shouldn't try to avoid using sh -c (or equivalent) to create a new, separate shell process for this purpose, because:

  • Once the shell has become myCommand, there's no shell waiting to run subsequent commands. sh -c 'echo $$; exec myCommand; foo would not be able to attempt to run foo after replacing itself with myCommand. Unless you're writing a script that runs this as its last command, you can't just use echo $$; exec myCommand in a shell where you are running other commands.
  • You cannot use a subshell for this. (echo $$; exec myCommand) may be syntactically nicer than sh -c 'echo $$; exec myCommand', but when you run $$ inside ( ), it gives the PID of the parent shell, not of the subshell itself. But it is the subshell's PID that will be the PID of the new command. Some shells provide their own non-portable mechanisms for finding the subshell's PID, which you could use for this. In particular, in Bash 4, (echo $BASHPID; exec myCommand) does work.

Finally, note that some shells will perform an optimization where they run a command as if by exec (i.e., they forgo forking first) when it is known that the shell will not need to do anything afterward. Some shells try to do this anytime it is the last command to be run, while others will only do it when there are no other commands before or after the command, and others will not do it at all. The effect is that if your forget to write exec and just use sh -c 'echo $$; myCommand' then it will sometimes give you the right PID on some systems with some shells. I recommend against ever relying on such behavior, and instead always including exec when that's what you need.

  • Before I can run myCommand, I need to set a number of environment variables in my bash script. Will these carry over to the environment in which the exec command is running? Commented May 11, 2018 at 15:41
  • Looks like my environment does carry over into the exec command. However, this approach does not work when myCommand starts other processes, which are the ones you need to work with; when I issue a kill -INT <pid> where pid was obtained this way, the signal does not reach the sub-processes started by myCommand, whereas if I run myCommand in the current session and Ctrl+C, the signals propagate correctly. Commented May 11, 2018 at 16:43
  • 1
    I tried this, but the pid of the myCommand process seems to be the pid output by echo $$ +1. Am I doing something wrong?
    – crobar
    Commented Aug 28, 2018 at 10:13
  • My command looks like this: sh -c 'echo $$; exec /usr/local/bin/mbdyn -f "input.file" -o "/path/to/outputdir" > "command_output.txt" 2>&1 &'
    – crobar
    Commented Aug 28, 2018 at 10:51
  • 3
    This is brilliant, but it's not working for me when I try to get the echoed value into a variable so I can actually use it later to kill the process, e.g. PID=$(sh -c 'echo $$; exec myCommand') just hangs, whereas if I remove the PID=$(...) wrapper it displays the PID and continues immediately!
    – Michael
    Commented Apr 16, 2020 at 0:22

Wrap the command in a small script

yourcommand &
echo $! >/path/to/pid.file
  • This is the super-simple answer that I was looking for. EXACTLY what I needed.
    – PfunnyGuy
    Commented Jul 27, 2022 at 17:20

I do not know of any simpler solution, but isn't using $! good enough? You can always assign the value to some other variable if you need it later, as said by others.

As a side note, instead of piping from ps you could use pgrep or pidof.


use exec from a bash script after registering the pid to a file:


suppose you have a script named "forever.sh" that you want to run with args p1,p2,p3

forever.sh sourcecode:


while [ 1 -lt 2 ] ; do
    logger "$0 running with parameters \"$@\""
    sleep 5

create a reaper.sh:


echo $$ > /var/run/$1.pid
exec "$@"

run forever.sh through reaper.sh:

./reaper.sh ./forever.sh p1 p2 p3 p4 &

forever.sh does nothing more than logging a line to syslog each 5 seconds

you now have the pid in /var/run/forever.sh.pid

cat /var/run/forever.sh.pid 

and forever.sh is running aok. syslog grep:

Nov 24 16:07:17 pinkpony cia: ./forever.sh running with parameters "p1 p2 p3 p4"

you can see it in the process table:

ps axuwww|grep 'forever.sh p1' |grep -v grep
root      5780  0.0  0.0   4148   624 pts/7    S    16:07   0:00 /bin/sh ./forever.sh p1 p2 p3 p4
  • 3
    oh, and the "oneliner": /bin/sh -c 'echo $$>/tmp/my.pid && exec program args' &
    – user237419
    Commented Nov 24, 2010 at 14:28
  • 1
    To properly preserve internal whitespace in the arguments, you should use exec "$@" instead of exec $*. Technically what you need to preserve is not whitespace but occurrences of the characters in the IFS shell parameter (which defaults to space, tab, and newline). Commented Nov 25, 2010 at 3:53
  • point taken. :)
    – user237419
    Commented Nov 25, 2010 at 7:54
  • Thank you, I did not know $$ parameter. It can be very useful.
    – rafalmag
    Commented Nov 27, 2010 at 12:12

In the bash shell an alternative to $! might be the jobs -p built-in. In some cases the ! in $! gets interpreted by the shell before (or instead of) the variable expansion, leading to unexpected results.

This, for example, won't work:

((yourcommand) & echo $! >/var/run/pidfile)

while this will:

((yourcommand) & jobs -p >/var/run/pidfile)

You can use something like:

$ myCommand ; pid=$!


$ myCommand && pid=$!

The two commands can be joints using ; or &&. In the second case, the pid will be set only if the first command succeeds. You can get the process id from $pid.

  • 3
    OP wants to get the PID so he can kill it later. ; and && require the original process to exit before the echo $! is executed.
    – user9517
    Commented Nov 24, 2010 at 8:58
  • Yes, you are right. This will give you the pid after myCommand has terminated.
    – Khaled
    Commented Nov 24, 2010 at 9:02
  • 8
    Referencing $! after && or ; will never give you the PID of the process started for the left-hand side of the command separator. $! is only set for processes launched asynchronously (e.g. usually with & but some shells also have other methods). Commented Nov 24, 2010 at 9:47
  • it is helpful, and why no one vote up excep me? good job though :)
    – temple
    Commented Sep 28, 2016 at 21:14

This is a bit of a hack-y answer and won't likely work for most people. It's also a huge security risk methinks, so don't do it unless you're sure you'll be safe and the inputs are sanitized and...well, you get the idea.

Compile the little C program here into a binary called start (or whatever you want), then run your program as ./start your-program-here arg0 arg1 arg2 ...

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(int argc, char **argv)
    if (argc >= 2)
        printf("%lu\n", (long unsigned) getpid());
        if (execvp(argv[1], &argv[1]) < 0)
            return 127;
    return 0;

Long story short, this will print the PID to stdout, then load your program into the process. It should still have the same PID.

  • How can I record the PID number returned by this code in a bash variable? For a reason unclear to me, stdout does not seem to be recorded in this example: RESULT="$(./start_and_get_pid.out echo yo)"; echo "$RESULT"
    – Tfb9
    Commented Nov 8, 2018 at 1:27
  • @Tfb9: I'd honestly recommend one of the other approaches; I wrote this 2+ years ago and it's by far one of the hackier/error-prone methods presented (or so I think).
    – tonysdg
    Commented Nov 8, 2018 at 1:30
  • Thanks for the note. I think I solved my problem with this sleep 10 & PID_IS=$!; echo $PID_IS
    – Tfb9
    Commented Nov 9, 2018 at 0:33

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