Answer by glenn jackman for What is the difference between $$ and $! when using /bin/bash -c

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

$ echo $$; bash -c 'echo $$'; echo $$687269876872

$! is the PID of the last backgrounded process

$ sleep 5 & echo $!; sleep 5 & echo $!; echo ===; pgrep -l sleep[1] 71187118[2] 71197119===7118 sleep7119 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.

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.