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Haskell试图在[String]中改变字符串

[英]Haskell trying to mutate string in [String]

movex [] a s = []    
movex (x:xs) a s
| elem a x = moveNow x a s
| otherwise = x : (movex xs a s)
where
  moveNow x a s
    | s == 'l' = moveNow2 x a
    where
        moveNow2 [] _ = []
        moveNow2 (x:y:xs) a
          | x == ' ' && y == a = a : x : moveNow2 (y:xs) a
          | otherwise = x : moveNow2 (y:xs) a

<- This is what I got right now < - 这就是我现在所得到的

I am trying to make a function that iterates through [string], finds the right string and then mutates it. 我正在尝试创建一个遍历[string]的函数,找到正确的字符串,然后将其变异。

given input 给定输入

func ["abc", "dfg"] f l -- move f in this list 1 space left --

expected output 预期产出

["abc", "fdg"]

Right now I am stuck at movex function that gives me error 现在我被困在movex函数,给我错误

Couldn't match expected type `Char' with actual type `[Char]'
In the first argument of `(:)', namely `x'
In the expression: x : (movex xs a s)

Direct solution to the error is to replace the line 直接解决错误就是更换线路

| elem a x = moveNow x a s

With

| elem a x = moveNow x a s : movex xs a s

Or, probably 或者,可能

| elem a x = moveNow x a s : xs

Depending on what you want to do after the first match: continue looking for certain character, or leave other strings untouched. 取决于您在第一场比赛后想要做的事情:继续寻找某个角色,或保持其他角色不变。

Your moveNow function has return type String , or [Char] , while movex has [String] , or [[Char]] , that's why compiler complains. 你的moveNow函数有返回类型String[Char] ,而movex[String][[Char]] ,这就是编译器抱怨的原因。

To avoid such problems(or fix them easier) consider writing explicit type signatures, like so: 为避免此类问题(或更容易修复),请考虑编写显式类型签名,如下所示:

movex :: [String]->String->String->[String]

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