了解Haskell类型签名 [英] Understanding Haskell Type Signatures

问题描述

  Prelude> ：t（$）
（$）::（a  - > b） - > a  - > b 
前奏> 
  

我应该如何解读（无双关语意思）？

一个半相似的结果也被证明令人费解：

  Prelude> ：t map 
 map ::（a  - > b） - > [a]  - > [b] 
前奏> 
  

解决方案

我将以图。 map 函数将操作应用于列表中的每个元素。如果我有

  add3 :: Int  - > Int 
 add3 x = x + 3 
  

然后我可以将它应用到 Int 使用 map ：

$ b

 > map add3 [1，2，3，4] 
 [4,5,6,7] 
  

所以，如果你看看类型签名

  map ::（a  - > b） - > [a]  - > [b] 
  

您会看到第一个参数是（a - > b），它只是一个函数，它接受 a 并返回一个 b 。第二个参数是 [a] ，它是 a 类型的值列表，返回类型 [b] ，类型为 b 的值列表。因此，在普通英语中， map 函数将一个函数应用于值列表中的每个元素，然后将这些值作为列表返回。

这就是 map a 高阶函数的含义，它将函数作为参数并对它进行操作。查看 map 的另一种方法是在类型签名中添加一些括号以使其成为

  map ::（a  - > b） - > （[a]  - > [b]）
  

所以你也可以把它看作一个函数它将函数从 a 转换为 b ，转换为 [a] 到 [b] 。

---

（$）有类型

 （$）::（a - > b） - > a  - > b 
  

用于像

 > add3 $ 1 + 1 
 5 
  

它所做的只是向在这种情况下 1 + 1 ，并将它传递给 left 上的函数，在这里 ADD3 。为什么这很重要？它有一个方便的固定性或运算符优先级，它使它等价于

 > add3（1 + 1）
  

所以无论如何在传递给剩下。这对于将几个函数链接在一起很有用：

 > add3 $ add3 $ add3 $ add3 $ 1 + 1 
  

比$ / $更好b

 > add3（add3（add3（add3（1 + 1））））
  

因为你没有关闭括号。

I am in the process of teaching myself Haskell and I was wondering about the following type signatures:

Prelude> :t ($)
($) :: (a -> b) -> a -> b
Prelude>

How should I interpret (no pun intended) that?

A semi-similar result is also proving to be puzzling:

Prelude> :t map
map :: (a -> b) -> [a] -> [b]
Prelude>

解决方案

I'll start with map. The map function applies an operation to every element in a list. If I had

add3 :: Int -> Int
add3 x = x + 3

Then I could apply this to a whole list of Ints using map:

> map add3 [1, 2, 3, 4]
[4, 5, 6, 7]

So if you look at the type signature

map :: (a -> b) -> [a] -> [b]

You'll see that the first argument is (a -> b), which is just a function that takes an a and returns a b. The second argument is [a], which is a list of values of type a, and the return type [b], a list of values of type b. So in plain english, the map function applies a function to each element in a list of values, then returns the those values as a list.

This is what makes map a higher order function, it takes a function as an argument and does stuff with it. Another way to look at map is to add some parentheses to the type signature to make it

map :: (a -> b) -> ([a] -> [b])

So you can also think of it as a function that transforms a function from a to b into a function from [a] to [b].

---

The function ($) has the type

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

And is used like

> add3 $ 1 + 1
5

All it does is take what's to the right, in this case 1 + 1, and passes it to the function on the left, here add3. Why is this important? It has a handy fixity, or operator precedence, that makes it equivalent to

> add3 (1 + 1)

So whatever to the right gets essentially wrapped in parentheses before being passed to the left. This just makes it useful for chaining several functions together:

> add3 $ add3 $ add3 $ add3 $ 1 + 1

is nicer than

> add3 (add3 (add3 (add3 (1 + 1))))

because you don't have to close parentheses.

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