CODEWITHSUNDEEP
perlmoose
skiphoppy
skiphoppy

Reputation: 102793

Why does Moose's builder take a string value?

Moose::Manual::Attributes states:

As an alternative to using a subroutine reference [for default], you can instead supply a builder method for your attribute: ... This has several advantages. First, it moves a chunk of code to its own named method, which improves readability and code organization.

So, your attribute could define a default thusly:

has attr => (
  is => 'ro',
  builder => 'subroutine'
);
sub subroutine {
  # figure out and return default value
}

I don't get why this has to be separate from default. Couldn't you just pass a reference to a named subroutine?

has attr => (
  is => 'ro',
  default => \&subroutine
);

And would that not be better programming practice, since you're guaranteed not to accidentally refer to a subroutine that doesn't exist? You'd be referring to the method with a logical reference instead of a symbolic reference.

Upvotes: 7

Views: 737

Answers (3)

Matt S Trout
Matt S Trout

Reputation:

Subclassing.

Builder specifies a method name to call, so

package Something;
use Moose;

extends 'YourClass';

sub subroutine { <some other default> }

would have Something::subroutine called for a builder of 'subroutine', whereas if you used the subref style then the YourClass::subroutine would be called instead because you've made a direct reference to the subroutine rather than letting it go through method dispatch.

Upvotes: 7

rjbs
rjbs

Reputation: 1327

When the builder is called, this happens:

$object->$builder

If builder is a string (let's say build_attr), then users can write their own build_attr method in a subclass and it will be called. This makes the default value extensible via a simple named method mechanism.

If it's a reference to a subroutine, the reference is taken in the original class's package, which means it can't be overriden the same way.

Upvotes: 14

Dave Rolsky
Dave Rolsky

Reputation: 4532

It's not a "symbolic" reference. The builder is a method name. That means it is inheritable and composable from a role. If you pass a subroutine reference, that reference must exist in the same package (or be fully qualified).

I'm pretty sure I explain this in the manual. Is it unclear?

Upvotes: 10

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