perl - 使用 Perl 的 Class::Struct 构建对象时如何引用该对象？

Question

我是面向对象 Perl 的新手，我必须在同一对象的另一个子程序中访问同一对象的成员变量。示例代码在这里:

use Class::Struct;

struct Breed =>
{
    name  => '$',
    cross => '$',
};

struct Cat =>
[
    name     => '$',
    kittens  => '@',
    markings => '%',
    breed    => 'Breed',
    breed2 => '$',

];

my $cat = Cat->new( name     => 'Socks',
                    kittens  => ['Monica', 'Kenneth'],
                    markings => { socks=>1, blaze=>"white" },
                    breed    => { name=>'short-hair', cross=>1 },
                    ** //breed2 => sub { return $cat->breed->name;}**

                  );

print "Once a cat called ", $cat->name, "\n";
**print "(which was a ", $cat->breed->name, ")\n";**
print "had two kittens: ", join(' and ', @{$cat->kittens}), "\n";

但我不确定如何在breed2 的子程序中使用$cat->breed->name ？有人可以帮我弄这个吗。

Answer 1

breed2中的问题是您试图引用尚 undefined variable 。看起来它是同名的，但它不是您正在创建的对象。这有点鸡和蛋的问题。

无论如何，我不太确定您是否想要在该插槽中使用这样的匿名子程序。你是
只是想缩短 $cat->breed->name至 $cat->breed2 ?您可以从 undef 开始在 breed2并在构造函数之后立即更改其值，因为那时您将拥有对该对象的引用。然而，即使你在那里放置了一个子程序，你也必须取消引用它:

my $cat = Cat->new( name     => 'Socks',
                    kittens  => ['Monica', 'Kenneth'],
                    markings => { socks=>1, blaze=>"white" },
                    breed    => { name=>'short-hair', cross=>1 },
                    breed2   => undef,

                  );
$cat->breed2( sub { $cat->breed->name } );

print "Once a cat called ", $cat->name, "\n";
print "(which was a ", $cat->breed2->(), ")\n";
print "had two kittens: ", join(' and ', @{$cat->kittens}), "\n";