On Wed, Aug 5, 2009 at 05:21, Dermot<paik...@googlemail.com> wrote: > 2009/8/5 Chas. Owens <chas.ow...@gmail.com>: >> On Tue, Aug 4, 2009 at 23:07, Dave Tang<d.t...@imb.uq.edu.au> wrote: >>> On Wed, 05 Aug 2009 12:55:22 +1000, Chas. Owens <chas.ow...@gmail.com> >>> wrote: >>> >>> snip > > Here's someone's recent experience of encountering prototypes: > > http://perlhacks.com/2009/06/what-is-wrong-with-this-picture.php > > Good luck, > Dp. >
Yes, calling a function with & at the beginning will bypass a prototype. So will calling it via a reference (which means prototypes mean nothing the OO Perl): #!/usr/bin/perl use strict; use warnings; #one $ means this is a unary function (i.e. it takes one scalar) sub foo ($) { print join("::", @_), "\n"; } eval "foo(1, 2); 1"; #this will fail because we are passing in two arguments print $@; my $coderef = \&foo; $coderef->(1, 2, 3); #but this succeeds no problem -- Chas. Owens wonkden.net The most important skill a programmer can have is the ability to read. -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/