Haskell：从类型中推导函数 [英] Haskell: Deducing function from type

问题描述

例如，函数的定义

  twoply ::（a  - > b，a  - > c） - > a  - > （b，c）
  

对于我来说显而易见（如果我不使用 undefined :: a ）。

然后我想出了以下内容：

 ¢:: a  - > （a  - > b） - > b 
¢= flip（$）
  

其中有趣的属性

 （¢）¢（$）:: a  - > （a  - > b） - > b 
  

这引出了我的问题。给定 = :: = 为与...具有相同类型的关系，那么语句 x = :: = xx（$）唯一定义 x 的类型？必须 x = :: =¢，还是存在其他可能的类型 x ？

我尝试从 x = :: = xx（$）反推以推导出 x :: a - > （a - > b） - >

对于 x = const ，xx（$）也是如此，它的类型为 a - > b - >一个。所以它不能唯一标识这个类型。

So I was playing around with Haskell today, thinking about autogeneration of function definitions given a type.

For example, the definition of the function

twoply :: (a -> b, a -> c) -> a -> (b, c)

is obvious to me given the type (if I rule out use of undefined :: a).

So then I came up with the following:

¢ :: a -> (a ->b) -> b
¢ = flip ($)

Which has the interesting property that

(¢) ¢ ($) :: a -> (a -> b) -> b

Which brings me to my question. Given the relation =::= for "has the same type as", does the statement x =::= x x ($) uniquely define the type of x? Must x =::= ¢, or does there exist another possible type for x?

I've tried to work backward from x =::= x x ($) to deduce x :: a -> (a -> b) -> b, but gotten bogged down.

解决方案

x =::= x x ($) is also true for x = const, which has the type a -> b -> a. So it does not uniquely identify the type.

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