in reply to Prototype mismatch: sub B::OP::size: none vs () at

It's nothing you can really help and can be ignored by you. It has to do with a function being declared with an empty prototype (e.g. sub foo () {...}) and another with no prototype (e.g. sub bar {...}) and then having the prototyped one become an alias to the unprototyped one:
sub foo () { return "foo\n" } sub bar { return "bar\n" } *foo = \&bar;

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart