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我可以使用基于 Moose 的对象将正则表达式传递给 isa() 吗？

发布于 2024-08-25 05:17:19 字数 250 浏览 9 评论 0原文

我可以在 Moose 中使用 isa 并以正则表达式作为参数吗？如果不可能，我可以使用 ->isa 以外的其他方法实现相同的效果吗？

好的，有以下类型 Animal::Giraffe ， Animal::Carnivore::Crocodile ，我想做 ->isa(/^Animal:: /)，我可以这样做吗？如果不能，我可以用什么来达到预期的效果？

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┾廆蒐ゝ 2024-09-01 05:17:19

这些相关类型应该都“扮演”相同的角色，动物。然后你可以这样写：

has 'animal' => (
    is       => 'ro',
    does     => 'Animal',
    required => 1,
);

现在你有了比正则表达式更可靠的东西来确保程序的一致性。

These related types should all "do" the same role, Animal. Then you can write:

has 'animal' => (
    is       => 'ro',
    does     => 'Animal',
    required => 1,
);

Now you have something much more reliable than a regex to ensure the consistency of your program.

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一张白纸 2024-09-01 05:17:19

Leon Timmermans 的答案接近我的建议，尽管我会使用 Moose::Util::TypeConstraints 中的糖

use Moose;
use Moose::Util::TypeConstraints;

subtype Animal => as Object => where { blessed $_ =~ /^Animal::/ };

has animal => ( is => 'rw', isa => 'Animal' );

Leon Timmermans' answer was close to what I'd suggest though I'd use the sugar from Moose::Util::TypeConstraints

use Moose;
use Moose::Util::TypeConstraints;

subtype Animal => as Object => where { blessed $_ =~ /^Animal::/ };

has animal => ( is => 'rw', isa => 'Animal' );

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颜漓半夏 2024-09-01 05:17:19

扩展 perigrin 的答案，以便如果该类在其超类中的任何位置都有 Animal::* ，而不仅仅是在其直接类名称中（例如，如果 Helper::Monkey code> isa Animal::Monkey)：

use Moose;
use Moose::Util::TypeConstraints;

subtype Animal => 
  as Object =>
  where { grep /^Animal::/, $_->meta->linearized_isa };

has animal => ( is => 'rw', isa => 'Animal' );

我认为 jrockway 使用角色的建议有很多优点，但如果你想这样做，你最好涵盖所有基础。

Extending perigrin's answer so that it will work if the class has an Animal::* anywhere in its superclasses, and not only in its immediate class name (for example if Helper::Monkey isa Animal::Monkey):

use Moose;
use Moose::Util::TypeConstraints;

subtype Animal => 
  as Object =>
  where { grep /^Animal::/, $_->meta->linearized_isa };

has animal => ( is => 'rw', isa => 'Animal' );

I think jrockway's suggestion to use a role instead has a lot of merit, but if you want to go this way instead, you might as well cover all of the bases.

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何处潇湘 2024-09-01 05:17:19

我认为这应该可以做到。

use Moose;
use Moose::Util::TypeConstraints;

my $animal = Moose::Meta::TypeConstraint->new(
    constraint => sub { $_[0] =~ /^Animal::/}
);

has animal => (is => 'rw', isa => $animal);

ETA：不过我同意 jrockway 的观点：除非你有令人信服的理由，否则你应该只使用继承。

I think this should do it.

use Moose;
use Moose::Util::TypeConstraints;

my $animal = Moose::Meta::TypeConstraint->new(
    constraint => sub { $_[0] =~ /^Animal::/}
);

has animal => (is => 'rw', isa => $animal);

ETA: I agree with jrockway though: unless you have a convincing reason otherwise, you should just use inheritance.

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