I like the twisted logic but the best part is the bare $ at the end. I had to run it through B::Deparse to see what was happening:

    $ perl -MO=Deparse  -pe '$_=$'

    LINE: while (defined($_ = <ARGV>)) {
        $_ = $;
    }
    continue {
        die "-p destination: $!\n" unless print $_;
    }
[download]

It shows that in this case $ gets interpreted as $; the subscript separator. However, this means that your code inserts an extra character in the file as shown by this:     perl -pe '$.-32or$_=$' file | cat -A

But I'm still not sure why perl inserts that semicolon. It doesn't in other cases:

    $ perl -MO=Deparse  -pe '$_=$.'

    LINE: while (defined($_ = <ARGV>)) {
        $_ = $.
    }
    continue {
        die "-p destination: $!\n" unless print $_;
    }
[download]

Anyone have an explanation for this last point?

Update: blakem's answer below is right. This last case is a perl 5.005 issue with B::Deparse.

--
John.

I seem to get the semicolon with your second case:

% perl5.6.1 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.;
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }

% perl5.8.0 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.;
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }
[download]

Update: Response to above update: with 5.00503 I get:

% perl5.00503 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }
[download]

-Blake

won't "32or" get interpreted as a single token?

No it won't.

or32 would be seen as an identifier, though.