Haskell中的旋转功能 [英] Rotate function in Haskell

问题描述

我想在Haskell中编写一个函数，该函数将由第二个参数给出的列表旋转第一个参数指示的位置数.使用模式匹配，实现递归函数

I want to write a function in Haskell that rotates the list given as the second argument by the number of positions indicated by the first argument. Using pattern matching, implement a recursive function

我编写了以下函数:

rotate :: Int -> [a] -> [a]
rotate 0 [y]= [y]
rotate x [y]= rotate((x-1) [tail [y] ++ head [y]])

，但是此函数始终会产生错误.有什么办法解决吗?该函数在运行时应执行以下操作:

but this function always produces a error. Is there any way to solve it? The function should do the following when it runs:

rotate 1 "abcdef"
"bcdefa"

推荐答案

[y] 并不意味着让 y 成为列表".这意味着此参数是一个包含一个名为 y 的元素的列表".您拥有正确的结构，但不需要 y 周围的括号.

[y] does not mean "let y be a list". It means "this argument is a list containing one element called y". You have the right structure, but you don't need the brackets around the y.

rotate :: Int -> [a] -> [a]
rotate 0 y = y
rotate x y = rotate (x-1) (tail y ++ [head y])

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