bash setsid nohup ./prog & -- $! not pointed at child process

Question

I was trying to get pid of process I ran with setsid and which ought to run in background like this:

test.sh:

#/bin/bash
setsid nohup ./my_program &
echo $!

if I run ./test.sh it will print a pid of my_program process and it's exactly what I need. But if run this commands one by one in my shell like this:

$ setsid nohup ./my_program &
$ echo $!

It will give me a pid of setsid command (or may be something else, but it almost all times gives me pid of my_program minus one).

What is happening here? Why results of commands I ran in terminal by myself differs from results of test.sh script?

Btw, may be you know some easy way of process which I started with setsid and which I need to run in background?

Maybe it has to do with setsid's forking behavior? It only forks if it is the process group leader; see [here](http://stackoverflow.com/a/9685973/1735215). — 0range, Apr 22 '16 at 14:31
To get the pid of a process executed via setsid, you may try [this](http://stackoverflow.com/a/19767312/1735215) — 0range, Apr 22 '16 at 14:57
@0range yeah, I think first link is exactly what I needed. Thanks a lot! Could you please write this as an answer so I can accept it? — PepeHands, Apr 22 '16 at 17:00
Please try to come up with a title specific enough that folks can know what your question is about (thus, if they're having their own problem with $!, whether answers to your question are also applicable and useful to them) without needing to click through and read it; I've tried to edit it to do that here. — Charles Duffy, Apr 22 '16 at 22:59

score 1 · Accepted Answer · edited May 23 '17 at 12:16

1

Repost of comments above as an answer:

This is because setsid only forks the current process if it is the process group leader. A detailed explanation can be found here.

To get the pid of a process executed via setsid, the approaches given here may be tried.

edited May 23 '17 at 12:16

Community

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answered Apr 22 '16 at 18:41

0range

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score 1 · Answer 2 · answered Apr 22 '16 at 22:45

setsid will call fork to ensure that it creates a new process group aswell as a new session, hence the resulting pid will not match the pid of setsid. The most clean work-around would be that my_program stores its pid into a file.

When you later want to send kill to my_program, you should check that the pid actually matches a program named my_program via /proc file system or calling the ps command with some magic code around it. (This is a very common method used by many daemons)

bash setsid nohup ./prog & -- $! not pointed at child process

2 Answers2