Moose：角色中的共享属性答案

【问题标题】：Moose: Share attribute in roleMoose：角色中的共享属性
【发布时间】：2015-02-20 11:14:47
【问题描述】：

我想在角色中声明一个属性，并且它的值在使用该角色的所有类实例之间共享。

我已经写了这个，但我认为这不是更好的方法：

package RealRessource;
use Moose;
use MooseX::ClassAttribute;
class_has '_real_ressource' => ( is => 'ro', isa => 'Int', lazy => 1,
                                 builder => '_build_real_ressource' );

sub _build_real_ressource {
    print "_build_real_ressource\n";
    return int(rand(100));
}
__PACKAGE__->meta->make_immutable;

package ShareRessource;
use Moose::Role;
has 'ressource' => ( is => 'ro', isa => 'Int', lazy => 1,
                     builder => '_build_ressource' );

sub _build_ressource {
    print "Build New Ressource\n";
    my $real_ressource = new RealRessource();
    return $real_ressource->_real_ressource;
}

package A;
use Moose;

with 'ShareRessource';

__PACKAGE__->meta->make_immutable;

package B;
use Moose;

with 'ShareRessource';

__PACKAGE__->meta->make_immutable;

package main;
use A;
use B;

my $a = new A();
my $b = new B();
print $a->ressource,$/;
print $b->ressource,$/;

结果是：

Build New Ressource
_build_real_ressource
28
Build New Ressource
28

【问题讨论】：

标签： perl moose

【解决方案1】：

此解决方案取消了额外的 RealResource 类。该角色被转换为一个类，以便它可以共享类属性。 MooseX::ABC 的使用是可选的，但我把它放在那里是为了强制它不能被实例化。

package ShareResource;
use Moose;
use MooseX::ABC;
use MooseX::ClassAttribute;
class_has 'resource' => (
    is      => 'ro',
    isa     => 'Int',
    lazy    => 1,
    builder => '_build_resource',
);
sub _build_resource {
    print "_build_resource\n";
    return int(rand(100));
}
__PACKAGE__->meta->make_immutable;

package A;
use Moose;
extends 'ShareResource';
__PACKAGE__->meta->make_immutable;

package B;
use Moose;
extends 'ShareResource';
__PACKAGE__->meta->make_immutable;

package main;
use A;
use B;

my $a = A->new();
my $b = B->new();
print $a->resource,$/;
print $b->resource,$/;

【讨论】：