ruby - 动态扩展现有方法或覆盖 ruby​​ 中的发送方法

Question

假设我们有 A、B、C 类。

A
 def self.inherited(sub)
   # meta programming goes here
   # take class that has just inherited class A
   # and for foo classes inject prepare_foo() as 
   # first line of method then run rest of the code
 end

 def prepare_foo
   # => prepare_foo() needed here
   # some code
 end

end

B < A
  def foo
    # some code
  end
end

C < A
  def foo
    # => prepare_foo() needed here
    # some code
  end
end

如您所见，我正在尝试foo_prepare()对每个foo()方法注入调用。

怎么可能呢？

此外，我一直在考虑以我会运行的方式覆盖send类，而不是让(super) 来执行其余的方法。class Afoo_preparesend

你们怎么看，解决这个问题的最佳方法是什么？

Answer 1

这是给你的解决方案。尽管它基于模块包含而不是从类继承，但我希望您仍然会发现它很有用。

module Parent
  def self.included(child)
    child.class_eval do
      def prepare_for_work
        puts "preparing to do some work"
      end
  
      # back up method's name
      alias_method :old_work, :work
  
      # replace the old method with a new version, which has 'prepare' injected
      def work
        prepare_for_work
        old_work
      end
    end
  end
end

class FirstChild
  def work
    puts "doing some work"
  end

  include Parent # include in the end of class, so that work method is already defined.
end

fc = FirstChild.new
fc.work
# >> preparing to do some work
# >> doing some work

Answer 2

我推荐Sergio 的解决方案（已接受）。这是我所做的符合我的需求。

class A
  def send(symbol,*args)
    # use array in case you want to extend method covrage
    prepare_foo() if [:foo].include? symbol
    __send__(symbol,*args)
  end
end

或者

class A
  alias_method :super_send, :send           

  def send(symbol,*args)
    prepare_foo() if [:foo].include? symbol
    super_send(symbol,*args)
  end
end

Answer 3

从 Ruby 2.0 开始，您可以使用 'prepend' 来简化 Sergio 的解决方案：

module Parent
  def work
    puts "preparing to do some work"
    super
  end
end

class FirstChild
  prepend Parent

  def work
    puts "doing some work"
  end
end

fc = FirstChild.new
fc.work

这允许模块在不需要 alias_method 的情况下覆盖类的方法。