perl -lane '$s+=$F[2]}{print $s/$.'

-sauoq
"My two cents aren't worth a dime.";

okay, i'll bite. why does this work?

I mean, using an explicit END block i understand...

perl -lane '$s+=$F[2]; END{print $s/$.;}'

But what is it about your "}{" that makes the print part of an implicit END block? Is this documented behavior? or an unintenional artifact of the way the parser deals with -n ?

Is this documented behavior? or an unintenional artifact of the way the parser deals with -n ?

Both. :-) (But you have to read the documentation creatively.) From perlrun:

-n   causes Perl to assume the following loop around your
     program, which makes it iterate over filename arguments
     somewhat like sed -n or awk:

       LINE:
         while (<>) {
             ...             # your program goes here
         }
[download]

When you put a '}{' in the middle of that it has the effect of ending the while loop and beginning another block (which is then finished by the old end of the while loop's block.)

I think I picked it up from someone's round of golf. I'd give credit if I knew to whom it was due.

-sauoq
"My two cents aren't worth a dime.";