Question

class A
  def a
    puts 'in #a'
  end
end

class B < A
  def a
    b()
  end
  def b
    # here i want to call A#a.
  end
end

Answer 1

class B < A

  alias :super_a :a

  def a
    b()
  end
  def b
    super_a()
  end
end

Answer 2

没有很好的方法可以做到这一点，但是你可以做A.instance_method(:a).bind(self).call，这会有效，但很难看。

你甚至可以在Object中定义自己的方法，在java中表现得像super：

class SuperProxy
  def initialize(obj)
    @obj = obj
  end

  def method_missing(meth, *args, &blk)
    @obj.class.superclass.instance_method(meth).bind(@obj).call(*args, &blk)
  end
end

class Object
  private
  def sup
    SuperProxy.new(self)
  end
end

class A
  def a
    puts "In A#a"
  end
end

class B<A
  def a
  end

  def b
    sup.a
  end
end
B.new.b # Prints in A#a

Answer 3

如果你没有明确需要从B＃b调用A＃a，而是需要从B＃a调用A＃a，这实际上是你通过B＃b做的事情（除非你'例子不够完整，无法证明你为什么要从B＃b打电话，你可以在B＃a中调用super，就像有时在初始化方法中做的一样。我知道这很明显，我只是想要澄清任何Ruby新手，你不需要别名（特别是这有时被称为“围绕别名”）。

class A
  def a
    # do stuff for A
  end
end

class B < A
  def a
    # do some stuff specific to B
    super
    # or use super() if you don't want super to pass on any args that method a might have had
    # super/super() can also be called first
    # it should be noted that some design patterns call for avoiding this construct
    # as it creates a tight coupling between the classes.  If you control both
    # classes, it's not as big a deal, but if the superclass is outside your control
    # it could change, w/o you knowing.  This is pretty much composition vs inheritance
  end
end

在ruby中调用超类中的另一个方法

3 个答案: