继承但不与超类共享且不共享命名空间的变量/常量

Question

有没有办法通过以下三个属性引入一些变量/常数？

a) 当它没有在自己的类中分配时，它会继承超类的值。

b) 它不继承超类以外的其他类的值（即使它们共享命名空间）。

c) 在自己的类中赋值时不会覆盖超类的值。

使用类实例变量，a) 不满足。

class A; @foo = :foo end
class B < A; @foo end # => nil  (Does not satisfy (a))
class A; class C; @foo end end # => nil (Satisfies (b))

class B < A; @foo = :bar end
class A; @foo end # => :foo  (Satisfies (c))

使用类变量，c) 不满足。

class A; @@foo = :foo end
class B < A; @@foo end # => :foo  (Satisfies (a))
class A; class C; @foo end end # => NameError (Satisfies (b))

class B < A; @@foo = :bar end
class A; @foo end # => :bar  (Does not satisfy (c))

使用常数，b) 不满足。

class A; Foo = :foo end
class B < A; Foo end # => :foo  (Satisfies (a))
class A; class C; Foo end end # => :foo (Does not satisfy (b))

class B < A; Foo = :bar end
class A; Foo end # => :foo  (Satisfies (c))

我想要这样的东西：

class A; something = :foo end
class B < A; something end # => :foo  (Satisfies (a))
class A; class C; something end end # => nil or Error (Satisfies (b))

class B < A; something = :bar end
class A; something end # => :foo  (Satisfies (c))

如果不能简单地通过分配和引用变量/常量来完成，那么有没有办法实现具有此属性的访问器方法？

Answer 1

您需要使用所需的特定属性创建自己的访问器类型。例如，

module InheritableProperty
  def property
    @property || superclass.property
  end
  def property=(value)
    @property = value
  end
end

class A
  extend InheritableProperty
end
class B < A
  extend InheritableProperty
  class C
    extend InheritableProperty
  end
end

A.property = 1
A.property # => 1
B.property # => 1
B::C.property # error

A.property = 1
B.property = 2
A.property # => 1
B.property # => 2
B::C.property # error

A.property = 1
B.property = 2
B::C.property = 3
A.property # => 1
B.property # => 2
B::C.property # => 3

Answer 2

按照 joshuanapoli 的建议，我决定这样做：

class A
  def self.foo; defined?(@foo) ? @foo : superclass.foo end
end

class A; @foo = :foo end
class B < A; foo end # => :foo
class A; class C; foo end end # => Error
class B < A; @foo = :bar end
class A; foo end # => :foo