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Re: How to create nested class in perl?

by Bloodnok (Vicar)
on Nov 13, 2009 at 22:19 UTC ( #807069=note: print w/ replies, xml ) Need Help??


in reply to How to create nested class in perl?

If your aim is to create a class visible only to the enclosing class, then something along the following lines might suffice:

# File: Frobnicate.pm use warnings; use strict; package Util; . . package Frobnicate::Bar; . . package Frobnicate::Foo; our @ISA = qw/Frobnicate/; . . package Frobnicate; . . 1;
Note that, within Frobnicate.pm, ...
  • All classes (packages) are visible to each other.
  • Only Frobnicate is accessible (& thus available as a superclass) externally.
  • Frobnicate::Foo subclasses (and thus has a 'relationship' with) Frobnicate.
  • Frobnicate::Bar has no 'relationship' with Frobnicate
Also, note that there is no need to use the subclass & utility classes where sub-classing behaviour is required.

Update:

The most apposite & absolutely correct) observations of both GrandFather & dsheroh have, between them, served to most ably demonstrate the...

  • Voracity of my sig here on the monastry
  • Correctness/validity of the (sometimes self-imposed) coding standards (&/or conventions) to which I have worked - my posting was purely experience based

A user level that continues to overstate my experience :-))


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Re^2: How to create nested class in perl?
by GrandFather (Cardinal) on Nov 14, 2009 at 02:56 UTC

    Given:

    use warnings; use strict; package Frobnicate::Bar; sub new { my ($class) = @_; return bless {}, $class; } sub sayWhat { my ($self) = @_; if (ref $self) { print "I'm a ", ref $self, " method\n"; } else { print "I'm a ", __PACKAGE__, " function\n"; } } package Frobnicate::Foo; our @ISA = qw/Frobnicate/; sub new { my ($class) = @_; return Frobnicate::new ($class); } sub sayWhat { my ($self) = @_; if (ref $self) { print "I'm a ", ref $self, " method\n"; } else { print "I'm a ", __PACKAGE__, " function\n"; } } package Frobnicate; sub new { my ($class) = @_; return bless {}, $class; } sub sayWhat { my ($self) = @_; if (ref $self) { print "I'm a ", ref $self, " method\n"; } else { print "I'm a ", __PACKAGE__, " function\n"; } } 1;

    consider:

    use strict; use warnings; use noname1; Frobnicate::Bar->new ()->sayWhat (); Frobnicate::Foo->new ()->sayWhat (); Frobnicate->new ()->sayWhat (); Frobnicate::Bar::sayWhat (); Frobnicate::Foo::sayWhat (); Frobnicate::sayWhat ();

    which prints:

    I'm a Frobnicate::Bar method I'm a Frobnicate::Foo method I'm a Frobnicate method I'm a Frobnicate::Bar function I'm a Frobnicate::Foo function I'm a Frobnicate function

    So, exactly what is hidden from the calling code exactly?

    btw, I don't condone the use of the same sub as a method and a function. Used here to reduce code peripheral to the main thesis.


    True laziness is hard work
Re^2: How to create nested class in perl?
by dsheroh (Parson) on Nov 14, 2009 at 11:48 UTC
    Only Frobnicate is accessible (& thus available as a superclass) externally.

    Incorrect. Any external code which does a use Frobnicate will get access to all packages/classes within Frobnicate.pm. The relationship of "file name == package name" is purely a matter of convention. It is not enforced in any way by perl.

    Let's create a module Foo.pm containing:

    package Bar; sub identify { print "I'm in package " . __PACKAGE__ . "\n"; } 1;
    and then write a program that says:
    #!/usr/bin/perl use Foo; Bar->identify(); Foo->identify();
    Running this program produces the output
    I'm in package Bar Can't locate object method "identify" via package "Foo" (perhaps you f +orgot to load "Foo"?) at...
    As you can see, Bar being named differently than the file it's in does not in any way reduce its visibility or accessibility to the outside world.

    But don't do that in production code, of course. The reason for the convention is because use Foo giving you access to package Bar is confusing as hell, especially if it doesn't give access to package Foo.

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