如何显示值的类型层次结构？

Question

在Perl5和Moose中,线性isa或线性化isa有助于理解类层次结构.

WHAT方法显示了值的具体类型:

> 42.WHAT
(Int)

我该如何表现出类似的东西

> 42.hypothetical-type-hierarchy
(Int) ? is (Cool) ? is (Any) ? is (Mu)
      ? does (Real) ? does (Numeric)

...可能还有每个消费角色的更多行？

---

编辑:具有两个角色的示例

class Beta {}
role Delta {}
role Gamma does Delta {}
role Eta {}
role Zeta does Eta {}
role Epsilon does Zeta {}
class Alpha is Beta does Gamma does Epsilon {}

# (Alpha) ? is (Beta)
#         ? does (Gamma) ? does (Delta)
#         ? does (Epsilon) ? does (Zeta) ? does (Eta)

my $ai = Alpha.new
$ai.^mro        # ((Alpha) (Beta) (Any) (Mu))

$ai.^roles      # ((Epsilon) (Zeta) (Eta) (Gamma) (Delta))
                # flat list, not two-element list of a tuple and triple?

Answer 1

您可以使用查询元对象

> 42.^mro
((Int) (Cool) (Any) (Mu))

其中,mro立场方法解析顺序和

> 42.^roles
((Real) (Numeric))

您可以控制通过副词返回的:local角色(从父类继承的角色 - 仅在类上可用)和:!transitive(排除通过其他角色组成的角色 - 在角色和类上都可用).

以下内容可以帮助您入门:

my $depth = 0;
for Alpha.^mro {
    say "is {.^name}";
    (sub {
        ++$depth;
        for @_ {
            say '  ' x $depth ~ "does {.^name}";
            &?ROUTINE(.^roles(:!transitive)); # recursive call of anon sub
        }
        --$depth;
    })(.^roles(:local, :!transitive));
}

鉴于您的示例代码稍作修改

role Delta {}
role Gamma does Delta {}
role Eta {}
role Zeta does Eta {}
role Epsilon does Zeta {}
class Beta does Gamma {}
class Alpha is Beta does Gamma does Epsilon {}

它产生输出

is Alpha
  does Epsilon
    does Zeta
      does Eta
  does Gamma
    does Delta
is Beta
  does Gamma
    does Delta
is Any
is Mu