如何定义此功能的类型配置文件？ [英] How to define the type profile for this function?

问题描述

 两次f x = f（f x）;我必须定义这个函数的类型配置文件。 
  

结果应该如下，但我不明白为什么。

 （'a  - >'a） - > 'a  - > 'a 
  

解决方案 > a） - > a - >一个是正确的答案。让我们把它分成几部分来找出原因。

1. 你的函数有两个参数， f 和 x ，因此签名将包含三个部分 - 比如 a - > c - > d


2. 这些参数中的第一个是一元函数 - 它使得 a =（a - > b） a 可以是任何类型，只要它在签名中只出现一次）并且签名看起来像（a - > ; b） - > c - > d


3. 结果两次与第一个参数的结果相同 - 这使得 d = b 和签名（a - > b） - > c - > b


4. f 第二个参数为两次作为它的参数 - 这使得 c = a 并且签名看起来像这样：（a - > b） - > a - > b


5. 两次应用于其自己的输出，这意味着 a = b - 这使得最终签名（a - > a） - > a - > a

I have to define the type profile of this function:

twice f x = f (f x);

The result should be the following, but I don't really get why.

('a -> 'a) -> 'a -> 'a

解决方案

(a -> a) -> a -> a is the right answer. Let's split it to pieces to find out why.

1. your function takes two arguments, f and x, so the signature will have three parts - say, a -> c -> d
2. first of these arguments is an unary function - that makes a = (a -> b) (remember that a can be any type, as long as it appears only once in the signature) and the signature look like (a -> b) -> c -> d
3. result of twice is same as the result of its first argument - that makes d = b and the signature (a -> b) -> c -> b
4. f takes second argument of twice as its argument - this makes c = a and the signature look like this: (a -> b) -> a -> b
5. twice is applied to its own output, which means that a = b - this makes the final signature (a -> a) -> a -> a

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