Ruby Koans:对类定义第2部分的显式范围

Question

我想澄清这篇原帖的一些内容.答案表明Ruby按此顺序搜索常量定义:

1. 封闭范围
2. 任何外部范围(重复到达顶级)
3. 包含的模块
4. 超(ES)
5. 宾语
6. 核心

那么澄清一下,在哪一步(1-6)是LEGS找到的常数的值legs_in_oyster？它来自超类Animal吗？是否MyAnimals忽略了类的范围,因为它不被视为封闭范围？这是由于显式的MyAnimals::Oyster类定义吗？

谢谢!只是想了解.这是代码:

 class Animal
   LEGS = 4
   def legs_in_animal
     LEGS
   end

   class NestedAnimal
     def legs_in_nested_animal
       LEGS
     end
   end
 end

 def test_nested_classes_inherit_constants_from_enclosing_classes
   assert_equal 4, Animal::NestedAnimal.new.legs_in_nested_animal
 end

 # ------------------------------------------------------------------

 class MyAnimals
   LEGS = 2

   class Bird < Animal
     def legs_in_bird
       LEGS
     end
   end
 end

 def test_who_wins_with_both_nested_and_inherited_constants
   assert_equal 2, MyAnimals::Bird.new.legs_in_bird
 end

 # QUESTION: Which has precedence: The constant in the lexical scope,
 # or the constant from the inheritance heirarachy?

 # ------------------------------------------------------------------

 class MyAnimals::Oyster < Animal
   def legs_in_oyster
     LEGS
   end
 end

 def test_who_wins_with_explicit_scoping_on_class_definition
   assert_equal 4, MyAnimals::Oyster.new.legs_in_oyster
 end

 # QUESTION: Now Which has precedence: The constant in the lexical
 # scope, or the constant from the inheritance heirarachy?  Why is it
 # different than the previous answer?
 end

Answer 1

我只是在同一个公案中思考同样的问题.我不是范围界定的专家,但以下简单的解释对我来说很有意义,也许它对你也有帮助.

当你定义MyAnimals::Oyster你仍然在全局范围内时,所以ruby不知道LEGS设置为2 的值,MyAnimals因为你实际上并不属于MyAnimals(有点违反直觉)的范围.

但是,如果要以Oyster这种方式定义,情况会有所不同:

class MyAnimals
  class Oyster < Animal
    def legs_in_oyster
      LEGS # => 2
    end
  end
end

不同之处在于,在上面的代码中,在您定义的时候Oyster,您已经进入了范围MyAnimals,因此ruby知道LEGS引用MyAnimals::LEGS(2)而不是Animal::LEGS(4).

仅供参考,我从以下网址获得了这一见解(在您链接的问题中引用):

• https://groups.google.com/forum/#!topic/comp.lang.ruby/t5rtDNol3P8