From: 120 <zen158...@zen.co.uk>
> I've looked at this:
>
> sub encrypt {
> my $self = shift;
> my $xx = $$self;
> #.. cut stuff I do understand
>
> return $self->SUPER::encrypt();
> }
>
> Could someone help me with the Perl to English here?
> I get that $self is shifting the arguement.
$self is set to the value of the first parameter of the subroutine
encrypt(). In case the encrypt() was called as a method, it will be
the object you called the method on
$obj->encrypt() -> $self = $obj;
> I think I get that $xx is a reference to $self?
No. $self was a reference to a scalar, $xx = $$self puts the value of
that scalar into $xx.
> I don't get at all: return $self->SUPER::encrypt();
>
> Is this assigning the $self to the results of a call to a class called
> SUPER's method encrypt?
No. This returns the result of calling the inherited version of the
encrypt() method on the object.
Jenda
===== je...@krynicky.cz === http://Jenda.Krynicky.cz =====
When it comes to wine, women and song, wizards are allowed
to get drunk and croon as much as they like.
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