使用foldl实现Haskell的take功能 [英] Implementing Haskell's `take` function using `foldl`

问题描述

使用foldl实现Haskell的take和drop函数.

Implementing Haskell's take and drop functions using foldl.

关于如何使用foldl ??

take x ls = foldl ???

drop x ls = foldl ???

我尝试了这些，但是显示错误:

i've tried these but it's showing errors:

myFunc :: Int -> [a] -> [a]
myFunc n list = foldl func [] list
    where 
    func x y | (length y) > n = x : y 
             | otherwise      = y

发生错误:

*** Expression : foldl func [] list
*** Term : func
*** Type : a -> [a] -> [a]
*** Does not match : [a] -> [a] -> [a]
*** Because : unification would give infinite type

推荐答案

我注意到您从未指定折叠必须在提供的列表上方.因此，一种可能满足您问题的 letter 的方法是:

I note that you never specified the fold had to be over the supplied list. So, one approach that meets the letter of your question, though probably not the spirit, is:

sillytake :: Int -> [a] -> [a]
sillytake n xs = foldl go (const []) [1..n] xs
  where go f _ (x:xs) = x : f xs
        go _ _ []     = []

sillydrop :: Int -> [a] -> [a]
sillydrop n xs = foldl go id [1..n] xs
  where go f _ (_:xs) = f xs
        go _ _ []     = []

这些都使用左折，但在数字列表[1..n]上-数字本身被忽略，并且仅使用列表的长度为其构建给定的自定义take n或drop n函数n.然后，此功能将应用于原始提供的列表xs.

These each use left folds, but over the list of numbers [1..n] -- the numbers themselves are ignored, and the list is just used for its length to build a custom take n or drop n function for the given n. This function is then applied to the original supplied list xs.

这些版本在无限列表上都可以正常工作:

These versions work fine on infinite lists:

> sillytake 5 $ sillydrop 5 $ [1..]
[6,7,8,9,10]

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