在Moose中创建类属性的最佳方式是什么?
问题描述:
我需要穆斯的类属性。现在我说:在Moose中创建类属性的最佳方式是什么?
#!/usr/bin/perl
use 5.010;
use strict;
use warnings;
use MooseX::Declare;
class User {
has id => (isa => "Str", is => 'ro', builder => '_get_id');
has name => (isa => "Str", is => 'ro');
has balance => (isa => "Num", is => 'rw', default => 0);
#FIXME: this should use a database
method _get_id {
state $id = 0; #I would like this to be a class attribute
return $id++;
}
}
my @users;
for my $name (qw/alice bob charlie/) {
push @users, User->new(name => $name);
};
for my $user (@users) {
print $user->name, " has an id of ", $user->id, "\n";
}
答
我发现MooseX :: ClassAttribute,但它看起来很丑。这是最干净的方式吗?
#!/usr/bin/perl
use 5.010;
use strict;
use warnings;
use MooseX::Declare;
class User {
use MooseX::ClassAttribute;
class_has id_pool => (isa => "Int", is => 'rw', default => 0);
has id => (isa => "Str", is => 'ro', builder => '_get_id');
has name => (isa => "Str", is => 'ro');
has balance => (isa => "Num", is => 'rw', default => 0);
#FIXME: this should use a database
method _get_id {
return __PACKAGE__->id_pool(__PACKAGE__->id_pool+1);
}
}
my @users;
for my $name (qw/alice bob charlie/) {
push @users, User->new(name => $name);
};
for my $user (@users) {
print $user->name, " has an id of ", $user->id, "\n";
}
答
老实说,我不认为这是所有麻烦的类属性的必要。对于只读类的属性,我只是使用一个返回常量的子。对于读写属性,在包一个简单的状态变量,通常的伎俩(我还没有碰到地方,我需要一些更复杂的任何方案。)
state $count = 0;
method _get_id {
return ++$count;
}
用词汇专用块可如果您需要5.10之前的兼容性,请使用此选项
这是正确的方法。你也可以创建一个类类(如果你喜欢它们在一个单独的包中),看看ClassAttribute在底层做什么 – castaway 2009-07-08 07:09:47
不是`use strict`和`use warnings`冗余于`use MooseX :: Declare` ? – 2009-12-09 23:41:36