perl - 如何在 Perl 和 Moose 中创建不可变对象(immutable对象)的循环图？

Question

这似乎是一个明显没有希望的情况，但是在 Perl 中创建不可变对象(immutable对象)的循环图有什么技巧吗？像这样的东西:

package Node;
use Moose;
has [qw/parent child/] => (is => 'ro', isa => 'Node');

package main;
my $a = Node->new;
my $b = Node->new(parent => $a);

现在，如果我想要 $a->child指向 $b ， 我能做些什么？

Answer 1

你可以玩延迟初始化的游戏:

package Node;
use Moose;

has parent => (
  is => 'ro',
  isa => 'Node',
  lazy => 1,
  init_arg => undef,
  builder => '_build_parent',
);

has _parent => (
  is => 'ro',
  init_arg => 'parent',
);

has child => (
  is => 'ro',
  isa => 'Node',
  lazy => 1,
  init_arg => undef,
  builder => '_build_child',
);

has _child => (
  is => 'ro',
  init_arg => 'child',
  predicate => undef,
);

has name => is => 'ro', isa => 'Str';

动态生成构建器和谓词:

BEGIN {
  for (qw/ parent child /) {
    no strict 'refs';

    my $squirreled = "_" . $_;

    *{"_build" . $squirreled} = sub {
      my($self) = @_;
      my $proto = $self->$squirreled;
      ref $proto eq "REF" ? $$proto : $proto;
    };

    *{"has" . $squirreled} = sub {
      my($self) = @_;
      defined $self->$squirreled;
    };
  }
}

这允许

my $a = Node->new(parent => \my $b, name => "A");
   $b = Node->new(child  =>     $a, name => "B");

for ($a, $b) {
  print $_->name, ":\n";
  if ($_->has_parent) {
    print "  - parent: ", $_->parent->name, "\n";
  }
  elsif ($_->has_child) {
    print "  - child: ", $_->child->name, "\n";
  }
}

它的输出是

A:
  - parent: B
B:
  - child: A

使用 η-conversion‎ 可以使代码更优雅，但 Moose 不会将参数传递给构建器方法。