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.
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