UpdateThis was already covered in depth by the rest of the thread, it would be safe for you to ignore it.

I think the reason you're geting the shell's pid is that the shell gets to the $$, sticks it's pid into there, and passes it's pid off to perl as the value of the argument to -i...

If I ask bash,

:~$ echo -i.$$
-i.27383
[download]

you see what's passed on to perl, and see that there is a literal numner ( 27383 here ) passed in ...

I think this was covered above, but i can't be sure.

HTH