Way to use Moo::Role to extend non-Moo class?

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

Fellow Monks;

I am trying to simplify some code here and would like to take a set of interfaces in one class and turn them into a Moo role.

I tries with and extends and neither one would work. It seems I can use base but that leads to a lot of problems. Any ideas?

I would like to do something like this (pseudocode):

package foo;
sub foo {
    print "foo\n";
}

package bar;
use Moo::Role;
extends 'foo';
[download]

I can't figure out how to accomplish this though. Is it even reasonably possible?

Comment on Way to use Moo::Role to extend non-Moo class? Download Code

Replies are listed 'Best First'.
Re: Way to use Moo::Role to extend non-Moo class? by Arunbear (Prior) on Nov 13, 2013 at 10:49 UTC
What is the purpose of your bar package? If you wanted to reuse methods from package foo without refactoring foo itself, you could use method forwarding: `package bar; use Moo::Role; has 'foo' => ( is => 'rwp', handles => [qw/foo/], # put other methods to be delegated here );` [download] Then any class that consumes this role will be able to call the methods of foo that you set up delegation for. If you were wanting to split up the methods into different roles, you could e.g. move the foo() method to its own role: `package Foo::Role; sub foo { print "foo\n"; }` [download] Then if bar wanted to 'extend' this with other methods: `package bar; use Moo::Role; with 'Foo::Role'; # ... other methods` [download]	[reply] [d/l] [select]
Re^2: Way to use Moo::Role to extend non-Moo class? by einhverfr (Friar) on Nov 13, 2013 at 11:07 UTC
In my case it won't be that simple, but it's very helpful in terms of thinking through the problem. Thanks.	[reply]
Re: Way to use Moo::Role to extend non-Moo class? by tobyink (Canon) on Nov 13, 2013 at 12:04 UTC
Roles don't "extend" classes. Subclasses extend classes. This will work: `package Foo { sub new { bless {}, shift; } sub foo { print "foo\n"; } } package Bar { use Moo; # NOT Moo::Role extends "Foo"; }` [download] ... but if the constructor for class `Foo` is any more complicated than the above, you'll need to write a `FOREIGNBUILDARGS` method for `Bar` to tell `Bar`'s constructor what argument need to be passed along to `Foo`'s constructor. `use Moops; class Cow :rw { has name => (default => 'Ermintrude') }; say Cow->new->name`	[reply] [d/l] [select]