我如何使用函数的monoid实例？

问题描述

GHCI告诉我，这个函数是一个Monoid类型的实例。 mconcat [id，id，id，id] 的类型是 Monoid a => a - >一个。但我预计它会是 a - > a 。

发生了什么？

解决方案

  instance Monoid b => Monoid（a  - > b）其中
 mempty _ = mempty 
 mappend fgx = fx`mappend` gx 
  

这是更通用的，因为它不需要同态（即 a - > a ）。为了得到你期待的实例，你可以将你的函数封装在 Endo 中：

  appEndo（mconcat [Endo id，Endo id，Endo id，Endo id]）
  

或

  appEndo $ mconcat $ fmap Endo [id，id，id，id] 
  

Today I tried to reduce a list of functions trough monoid typeclass but the resulting function expects its argument to be an instance of Monoid for some reason.

GHCI tells me that the type of mconcat [id, id, id, id] is Monoid a => a -> a. Yet I would expect it to be a -> a.

What is happening?

解决方案

You're using this instance:

instance Monoid b => Monoid (a -> b) where
    mempty _ = mempty
    mappend f g x = f x `mappend` g x

which is more general because it doesn't require endomorphisms (i.e. a -> a). To get the instance you were expecting, you can wrap your functions in Endo:

appEndo (mconcat [Endo id, Endo id, Endo id, Endo id])

or

appEndo $ mconcat $ fmap Endo [id, id, id, id]

这篇关于我如何使用函数的monoid实例？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！