What is the difference between $$ and $! when using /bin/bash -c

Question

I am trying to understand the "oneliner" in this comment by @sysfault:

/bin/sh -c 'echo $$>/tmp/my.pid && exec program args' &

I tried the following variations...

Different $$ inside and out:

/bin/bash -c 'echo $$ && exec sleep 5' ; echo $$ 14408 29700 

Same pid (but in background):

$ /bin/bash -c 'echo $$ && exec sleep 5' & echo $! [1] 8309 8309 $ 8309 [ 1 ]+ Done /bin/bash -c 'echo $$ && exec sleep 5' 

Different pid for $$ & $! :

/bin/bash -c 'echo $$ && exec sleep 5 & echo $!' 6504 6503 

Kill did not work: /bin/bash -c 'echo $$ && (sleep 15 ; echo done)'
19063 Killed $ done (echo showed up after the kill)

Same pid with access to the pid of the foreground process:

$ /bin/bash -c 'echo > tmp.pid ; sleep 10 ; echo $$ ; echo done ; echo $$;' 28415 (instantly) 28415 (10 seconds later) done 28415 $ cat tmp.pid 28415 

With this command I was able to kill the pid in another window before the sleep finished - preventing the last three echos from happening:

$ /bin/bash -c 'echo $$ ; sleep 30 ; echo $$ ; echo done ; echo $$;' 23236 Killed 

Context: Ultimately I want to start bash processes via php exec(), and wait for the result before continuing and save the pid so i can clean up in case the manual cancel button is pressed. There are 2 expensive processes i will deal with: one rsync and one git commit & push.

What is the best way to reliably get the pid of a foreground bash script that includes multiple commands (with ; > | or &&) ?

Answer 1

$$ is the PID of the current shell. When you use $$ in the bash -c script body, you get the pid of that spawned shell

$ echo $$; bash -c 'echo $$'; echo $$ 6872 6987 6872 

$! is the PID of the last backgrounded process

$ sleep 5 & echo $!; sleep 5 & echo $!; echo ===; pgrep -l sleep [1] 7118 7118 [2] 7119 7119 === 7118 sleep 7119 sleep 

When you do bash -c 'echo $$' & echo $!, you get the same pid printed twice because, for your current shell, $! is the pid of the backgrounded bash process, and inside the script body, $$ is the current pid of that same backgrounded bash process.

When you do /bin/bash -c 'echo $$ && exec sleep 5 & echo $!', you get two different pids: $$ is the pid of the bash process and $! is the pid of the sleep process.

• I find this odd because you're using exec so it should be the same PID (the sleep process replacing the bash process)
• However you're using exec ... & so bash must first fork a subshell to run the backgrounded task, and then exec replaces that subshell with sleep.

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The answer to your actual question is best answered by a php person, which I am not. You should flag this question to a moderator and request it be moved to Stack Overflow where more php people are bound to hang out.