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: - 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... - When I use my program to show pid between parenthesis, the pid it shows is different, is it right? - Is `$$` something like macro? Can you help me?

Accepted Answer

`$$` 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
```

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

Solution

Problem