在Perl5和Moose中,线性isa或线性化isa有助于理解类层次结构。
方法WHAT显示值的具体类型:
> 42.WHAT
(Int)
我如何显示类似
> 42.hypothetical-type-hierarchy
(Int) ┬ is (Cool) ─ is (Any) ─ is (Mu)
└ does (Real) ─ does (Numeric)
……可能需要为每个角色扮演更多角色?
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‽
最佳答案
您可以使用以下命令查询元对象
> 42.^mro
((Int) (Cool) (Any) (Mu))
其中
mro
代表方法解析顺序,而> 42.^roles
((Real) (Numeric))
您可以控制通过副词
:local
(不包括从父类继承的角色-仅在类中可用)和:!transitive
(不包括通过另一个角色组成的角色-在角色和类中均可用)返回的角色。以下内容将帮助您入门:
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));
}
给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
关于raku - 您如何显示值的类型层次结构?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/44255424/