find the package name of an inherited object from inside the parent

tomgracey has asked for the wisdom of the Perl Monks concerning the following question:

Hi All

I am sure this is a really simple question but cant figure out how to search for it (I'm not the sharpest stick in the box with Perl anyway tbh). This is a general Perl question, but I am using Moose so a Moose solution would be ok. Suppose I have 2 classes:

package Parent;

use Moose;

sub who_am_i{
   return __PACKAGE__;
}


package Child;

extends 'Parent';

... some specific Child methods ...
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Now if I do

my $child = Child->new;
print $child->who_am_i;
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I get Parent not Child. (At least this is what seems to happen from my experimentation. Unless I am doing something wrong!)

Is there a way the who_am_i method can be placed in the parent class, but return the package name of the actual initiated class (the child) ?

Sorry for such a basic question!

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Re: find the package name of an inherited object from inside the parent by Corion (Patriarch) on Jul 07, 2013 at 08:33 UTC
`__PACKAGE__` is a constant like `__FILE__` and `__LINE__`. What you might want is ref, which returns the class a reference is blessed into. Just be advised that outside of debugging, there is very little you should do based on the class name.	[reply] [d/l] [select]
Re^2: find the package name of an inherited object from inside the parent by tomgracey (Scribe) on Jul 07, 2013 at 09:27 UTC
Oh yeah! `ref($self)`! That really was a stupid question, apologies...	[reply] [d/l]
Re: find the package name of an inherited object from inside the parent by moritz (Cardinal) on Jul 07, 2013 at 09:15 UTC
The Mooseish way to get the dynamic class name is `$obj->meta->name`: `use 5.010; package Parent; use Moose; sub who_am_I { shift->meta->name }; package Child; use Moose; extends 'Parent'; package main; say Child->new->who_am_I; __END__ Child` [download] `$obj->meta` returns the metaclass instance for the object's class, and `->name` returns the name (surprise, surprise). (excursion: In Perl 6 it works basically the same, except that there is a shortcut for calling methods on the metaclass: `say 1.^name` says `Int`). Perl 6 - the future is here, just unevenly distributed	[reply] [d/l] [select]