为什么Haskell的“flip id”有这种类型？

问题描述

我很好奇表达式

id 函数有这样的类型：

  id :: a  - > a 
  

当您将 a 由 a - > b ：

  id ::（a  - > b） - > （a  - > b）
  

由于currying，它与以下相同：

  id ::（a  - > b） - > a  - > b 
  

现在将翻转 get：

  flip id :: a  - > （a  - > b） - > b 
  

id（+）实例是：

  id ::（Num a）=> （a  - > a） - > （a  - > a）
  

现在 flip id 给你：

  flip id ::（Num a）=> a  - > （a  - > a） - > a 
  

注意：这也显示了（$）与 id 相同，仅限于更受限制的类型：

 （$）::（a  - > b） - > a  - > b 
（$）fx = fx 
  - 非点：$ b​​ $ b（$）f = f 
  - 因此：
（$）= id 
  

I'm curious about the expression flip id (It's not homework: I found it in the getOpt documentation).

I wonder why it has this type:

Prelude> :t (flip id)
(flip id) :: b -> (b -> c) -> c

For example, (flip id) 5 (+6) gives 11.

I know why id (+6) 5 gives 11, but I don't "get" the flip id thing.

I tried to figure this out myself using pen and paper but couldn't. Could anybody please explain this to me? I mean, how does flip id come to have the type b -> (b -> c) -> c ?

解决方案

The id function has this type:

id :: a -> a

You get an instance of this type, when you replace a by a -> b:

id :: (a -> b) -> (a -> b)

which, because of currying, is the same as:

id :: (a -> b) -> a -> b

Now apply flip to this and you get:

flip id :: a -> (a -> b) -> b

In the case of id (+) the instance is:

id :: (Num a) => (a -> a) -> (a -> a)

Now flip id gives you:

flip id :: (Num a) => a -> (a -> a) -> a

Side note: This also shows you how ($) is the same as id, just with a more restricted type:

($) :: (a -> b) -> a -> b
($) f x = f x
-- unpoint:
($) f   = f
-- hence:
($)     = id

这篇关于为什么Haskell的“flip id”有这种类型？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！