函数获取一个Int并返回一个列表 [英] function that gets an Int and returns a list

问题描述

 <$ c在Haskell中我需要创建各种类型的练习。第一个类型称为Finite，定义如下： $ c> type Finite a = [a] 
  

然后我需要返回一个已定义的单例像这样

  singleF :: a  - >有限的
  

所以我实现了它：

 单身n = [n] 
  

创建另一个类型

  type Enumeration a = Int  - >有限的
  

然后我需要重新实现singleton函数

  singleE :: a  - >枚举a 
  

在我的理解类型 Enumeration 是一个从Int到< a 类型列表的函数的同义词，但我无法理解我能够如何实现它。

通过练习（前一种类型有限也被称为桶）：枚举是无限序列的有限桶，由自然数索引。 code>。

而函数single：为简单起见，我建议您将唯一的项目放在存储区0 中， m认为int是枚举中桶的索引

解决方案

关闭我的头顶：

  singleE :: a  - >枚举a 
 singleE a 0 = singleF a 
 singleE _ _ = [] 
 $ b $ main :: IO（）
 main = do 
 let s = singleE'a'
 print $ s 0 
 print $ s 5 
  

给出

 a

  

singleE给你一个函数，它接受一个I​​nt并返回一个Finite。如果你传递0，你会得到一个单一元素的Finite，否则是空的。

I have an exercise in Haskell where I need to create various types.The first type is called Finite which is defined like this:

type Finite a = [a]

and then I need to return a singleton which is defined like this

singleF :: a -> Finite a

so I implemented it like so:

single n = [n]

Then later I create another type

type Enumeration a = Int -> Finite a

then I need to reimplement the singleton function

singleE :: a -> Enumeration a

In my understanding the type Enumeration is a synonym for a function from an Int to a list of type a, but I can't understand how exactly I can implement that.

From the exercise (the previous type 'Finite' is also referred to as a 'bucket'): An enumeration is an infinite sequence of finite buckets, indexed by natural numbers.

And the function single : I suggest for simplicity that you put the sole item in bucket 0, So I'm thinking that the int is the index of the bucket in the enumeration

解决方案

Off the top of my head:

singleE :: a -> Enumeration a
singleE a 0 = singleF a
singleE _ _ = []

main :: IO ()
main = do
  let s=singleE 'a'
  print $ s 0
  print $ s 5

Gives

"a"
""

singleE gives you a function that takes an Int and returns a Finite. If you pass 0, you get a Finite with the single element, otherwise an empty one.

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