有没有办法将无限和有限的列表分开？ [英] Is there any way to separate infinite and finite lists?

问题描述

例如，我正在为列表编写一些函数，我想使用长度函数

  foo :: [a] - > Bool 
 foo xs = length xs == 100 
  

或者我应该总是考虑无限列表并使用类似这样的内容

  foo :: [a]  - > Bool 
 foo xs =长度（取101 xs）== 100 
  

代替长度直接吗？

如果haskell具有FiniteList类型，那么length和foo将会是

  length :: FiniteList a  - > Int 
 foo :: FiniteList a  - > Bool 
  

解决方案

length 遍历整个列表，但要确定列表是否具有特定长度 n ，则只需查看第一个 n 元素。

使用取的想法可行。或者
，你可以这样写一个 lengthIs 函数：

   - 假设n> = 0 
 lengthIs 0 [] = True 
 lengthIs 0 _ = False 
 lengthIs n [] = False 
 lengthIs n（x：xs）= lengthIs（n-1）xs 
  

您可以使用相同的思路编写 lengthIsAtLeast 和 lengthIsAtMost 变体。

For example, I am writing some function for lists and I want to use length function

foo :: [a] -> Bool
foo xs = length xs == 100

How can someone understand could this function be used with infinite lists or not?

Or should I always think about infinite lists and use something like this

foo :: [a] -> Bool
foo xs = length (take 101 xs) == 100

instead of using length directly?

What if haskell would have FiniteList type, so length and foo would be

length :: FiniteList a -> Int
foo :: FiniteList a -> Bool

解决方案

length traverses the entire list, but to determine if a list has a particular length n you only need to look at the first n elements.

Your idea of using take will work. Alternatively you can write a lengthIs function like this:

-- assume n >= 0
lengthIs 0 [] = True
lengthIs 0 _  = False
lengthIs n [] = False
lengthIs n (x:xs) = lengthIs (n-1) xs

You can use the same idea to write the lengthIsAtLeast and lengthIsAtMost variants.

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