Ruby 从父类(super class)方法让步到子类方法

Question

在 Ruby 中，是否可以在不影响子类代码的情况下从父类(super class)方法中产生子类方法？我试图避免在子类中调用 super。

class SuperClass
  def do_something
    puts "getting ready..."
    # how to then yield back to the subclass do_something method?
    puts "done."
  end
end

class SubClass < SuperClass
  def do_something
    # how to first execute the superclass do_something method?
    puts "doing something ..."
  end
end

期望的功能是专门调用 SubClass.do_something 并接收以下输出:

“准备……”

“做某事……”

“完成。”

---

编辑:

也许真正的问题是:如何干燥下面的代码，从所有子类中删除对 self.get_ready 和 self.finish_up 的调用，使用任何 ruby​​ meta - 使这些类保持干燥的编程技术:

class SuperClass
  def self.get_ready
    puts "getting ready ..."
  end

  def self.finish_up
    puts "done."
  end
end

class SubClassA < SuperClass
  def self.do_something
    self.get_ready
    puts "doing something ..."
    self.finish_up
  end
end

class SubClassB < SuperClass
  def self.do_something
    self.get_ready
    puts "doing something else, a little differently..."
    self.finish_up
  end
end

class SubClassC < SuperClass
  def self.do_something
    self.get_ready
    puts "doing something crazy..."
    self.finish_up
  end
end

Answer 1

在Practical Object-Oriented Design in Ruby ，Sandi Metz 建议您克服调用 super 的方法是在父类(super class)中定义一个“ Hook ”方法，唯一的工作就是被子方法覆盖。在你的情况下，你可以这样做:

class SuperClass
  def self.do_something
    puts "getting ready..."
    do_something_child
    puts "done."
  end

  def self.do_something_child; end
end

class SubClass < SuperClass
  def self.do_something_child
    puts "doing something ..."
  end
end

SubClass.do_something