Perl 在Moose中创建类属性的最佳方法是什么?
我需要Moose中的class属性。现在我说:Perl 在Moose中创建类属性的最佳方法是什么?,perl,moose,Perl,Moose,我需要Moose中的class属性。现在我说: #!/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');
#!/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";
}
我找到了MooseX::ClassAttribute,但它看起来很难看。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";
}
老实说,我认为没有必要为类属性带来那么多麻烦。对于只读类属性,我只使用返回常量的子类。对于读写属性,包中的一个简单状态变量通常起作用(我还没有遇到需要更复杂的情况)
如果需要5.10之前的兼容性,可以使用带有词法的私有块 啊,但是id_池不一定是只读的,我想要一个真正的类属性,它有一个访问器。当我决定将ID池移动到数据库时,我不想重写代码。如果它是一个访问器,我只需要更改访问器的工作方式,使用一个状态变量,我需要更改对它的所有引用,或者使它成为一个绑定标量。这是正确的方法。如果你喜欢在一个单独的包中创建一个类,你也可以创建一个类(实际上),看看ClassAttribute在这个包下做了什么,不是
使用strict
和使用warnings
与使用MooseX::Declare
?@Robert P也许,我写这篇文章的时候才刚刚开始使用MooseX::Declare。注意,从0.11开始,MooseX::ClassAttribute现在可以很好地处理角色。谢谢大卫·罗尔斯基!!Declare现在已被弃用。(他一边安装一边说。现在怎么办?)
state $count = 0;
method _get_id {
return ++$count;
}