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How do you get the PID in bash?

January 27th, 2014
tech, bash  [html]
If you ask how to get the PID in bash someone will tell you:
The variable $$ contains the PID.
This is usually correct, but not always. Sometimes it can be incorrect, but in a way that's actually closer to what you wanted.

The basic problem is that many bash constructs run in a subshell. This means the shell will fork itself, running your function in a separate process. This allows simple concurrency, but it also means a new process id. For example:

    $ function a { echo $$; }
    $ function b { echo $BASHPID; }
Function a uses $$ while b uses $BASHPID. When run as normal functions they both give the PID of bash running them:
    $ a
    $ b
When run in the background, however, they diverge:
    $ a &
    $ b &
To run a function in the background bash uses a multiprocess model, forking a subshell for each one, so we should expect both a & and b & to give new PIDs. In order to make this process faster, though, and because POSIX says so, bash doesn't reinitialize itself when creating a subshell. This means $$, which was computed once when bash started, continues to have its original value. $BASHPID always looks up the current PID of the executing process, however, so it does change when you make a subshell.

While this definition of $$ is frustrating if you actually want to get the current process id, for example to let each background function have free reign of /tmp/PID, in many cases this definition is actually helpful. There are many simple constructs in bash that require subshells, and it's convenient that $$ stays consistent. For example:

    $ ls $SOME_PLACE | while read fname ; do
        some_command $fname >> /tmp/$$.output
    $ other_command /tmp/$$.output
Because pipes run in a subshell if $$ were defined like $BASHPID all of these some_command outputs would build up in some random file in /tmp and other_command would look in a different file. Instead $$ happens to mean the right thing, and everything works.

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