[bash] Why is $$ returning the same id as the parent process?

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

"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?

If you were asking how to get the PID of a known command it would resemble something like this:

If you had issued the command below #The command issued was ***

dd if=/dev/diskx of=/dev/disky

Then you would use:

PIDs=$(ps | grep dd | grep if | cut -b 1-5)

What happens here is it pipes all needed unique characters to a field and that field can be echoed using

echo $PIDs

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

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

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.

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.