在动态语言中实现访问者模式的首选方法？

As an exercise for myself, I was translating a sample program into various languages. Starting in C#, I had a visitor-pattern interface like so:

作为我自己的练习,我正在将一个示例程序翻译成各种语言。从C#开始,我有一个访问者模式界面,如下所示:

interface Visitor
    {
        void Accept(Bedroom x);
        void Accept(Bathroom x);
        void Accept(Kitchen x);
        void Accept(LivingRoom x);
    }

Moving to ruby (or python, or whatever), I no longer get dispatch from the compiler based on type. Should I do the dispatch in the visitor itself?

转移到ruby(或python,或其他),我不再根据类型从编译器调度。我应该在访客本身进行调度吗?

class Cleaner

  def accept(x)
    acceptBedroom(x) if Bedroom === x
    acceptBathroom(x) if Bathroom === x
    acceptKitchen(x) if Kitchen===x
    acceptLivingRoom(x) if LivingRoom===x
  end

  ...

Or should I do the dispatch in the different specializaions of the room:

或者我应该在房间的不同专业中进行调度:

class Bathroom<Room
  def initialize(name)
    super(name)
  end
  def accept(visitor)
    visitor.acceptBathroom(self)
  end
end

Or is there another preferred idiom that is used in dynamic languages?

或者是否有另一种在动态语言中使用的首选习语?

2 个解决方案

#1

My recommendation is to use the former approach. There are pluses and minuses for each approach. The former is harder to maintain as the number of Room types grows; the latter is harder as the number of Cleaner types grows.

我的建议是使用前一种方法。每种方法都有优点和缺点。随着房间类型数量的增加,前者难以维持;随着清洁剂类型的数量增加,后者更难。

In Ruby, you could try

在Ruby中,你可以试试

def accept(x)
  send "accept#{x.class}".to_sym, x
end

PS: not all dynamically typed languages are unable to do dispatch based on type; some can infer type, or failing that, can used forced casting to pick the proper method among the overloaded options.

PS:并非所有动态类型的语言都无法根据类型进行调度;有些人可以推断类型,或者失败,可以使用强制转换在重载选项中选择正确的方法。

#2

I would go with the second version. The first one looks like the kind of code smell that Visitor is supposed to solve: long if-else-if or switch-case statements.

我会选择第二个版本。第一个看起来像访问者应该解决的那种代码气味:long if-else-if或switch-case语句。

#1