Why is $$ returning the same id as the parent process?

Question

I have problem with Bash, and I don't know why.
Under shell, I enter:

echo $$    ## print 2433
(echo $$)  ## also print 2433
(./getpid) ## print 2602

Where getpid is a C program to get current pid, like:

   int main() {
    printf("%d", (int)getpid());
    return 0;
   }

What confuses me is that:

1. I think "(command)" is a sub-process (am i right?), and i think its pid should be different with its parent pid, but they are the same, why...
2. When I use my program to show pid between parenthesis, the pid it shows is different, is it right?
3. Is $$ something like macro?

Can you help me?

Solution 1

$$ is defined to return the process ID of the parent in a subshell; from the man page under "Special Parameters":

$ Expands to the process ID of the shell. In a () subshell, it expands to the process ID of the current shell, not the subshell.

In bash 4, you can get the process ID of the child with BASHPID.

~ $ echo $$
17601
~ $ ( echo $$; echo $BASHPID )
17601
17634

Solution 2

You can use one of the following.

• $! is the PID of the last backgrounded process.
• kill -0 $PID checks whether it's still running.
• $$ is the PID of the current shell.

Solution 3

1. Parentheses invoke a subshell in Bash. Since it's only a subshell it might have the same PID - depends on implementation.
2. The C program you invoke is a separate process, which has its own unique PID - doesn't matter if it's in a subshell or not.
3. $$ is an alias in Bash to the current script PID. See differences between $$ and $BASHPID here, and right above that the additional variable $BASH_SUBSHELL which contains the nesting level.

Solution 4

Try getppid() if you want your C program to print your shell's PID.

Solution 5

this one univesal way to get correct pid

pid=$(cut -d' ' -f4 < /proc/self/stat)

same nice worked for sub

SUB(){
    pid=$(cut -d' ' -f4 < /proc/self/stat)
    echo "$$ != $pid"
}

echo "pid = $$"

(SUB)

check output

pid = 8099
8099 != 8100

View more solutions

Comments

[chepner]

Note that getpid would show a different process ID even if it weren't run in a subshell.

[Ben]

@Marian echo $$ $BASHPID ; ( echo $$ $BASHPID ) demonstrates that it does. Round brackets create a subshell. The statements may change variable values, and the parent shell must not see those changes. This is implemented as a fork() operation.

[Martin Bouladour]

"parent" is a bit misleading (at least it was to me), it's actually the "top level" shell. For instance : echo $$; (echo $$; (echo $$)) echoes the same pid three times

[chepner]

Right; I should have said that the value is inherited from a parent shell (which inherited its value from its parent, etc). The top level shell sets it initially, rather than inheriting from its (non-shell) parent process.

[Alexander Mills]

$ Expands to the process ID of the shell does it tho? echo $ just echoes the literal $.

[chepner]

@AlexanderMills Well, yes; $ alone is not a parameter expansion. The man page is referring to the name of the special parameter, which is $; it's not claiming that $ alone expands.

[Alexander Mills]

Ok I honestly have no idea what that means, but echo $BASHPID works in bash 4 and 5 (but not version 3.2.57 on MacOS)

[chepner]

All parameter expansions start with $, but $ is also the name of one of the special parameters. $, #, @, *, etc are some of the special parameters; $$, $#, $@, $*, etc are the expressions that expand to the value of each.

[Martin Dorey]

Nice idea, but won't that get you the pid of the fork the shell has run to capture the output of cut? If I run it twice, once with echo $(...) and once without, then I get different answers.

[Isaac Freeman]

Shouldn't the second bullet be kill -0 $! if we're talking about background processes? PID isn't set to anything by default.

[joseluisq]

Or if you use Rust just try nix::unistd::getppid() which will get the pid of current processes' parent - docs.rs/nix/0.18.0/nix/unistd/fn.getppid.html