Question

我想要List中一个元素的值，所以我做了

changePoint :: (Int,Int) -> [[[Char]]] -> [[Char]]

changePoint (x,y) maze = let 

          linelement = maze !! (y-1) 

-- value ["X"," "," ",...," ","X"]

          rowelement = chunksOf 1 $ head linelement 
-- type:[[a]]; value ["X"," "," "," "," "," "," "," "," "," "," "," "," ","X"]

          l = length rowelement
          list = take (x-1) rowelement

--          in take (x-1) rowelement  -- ["X"]  

           in take (x-1) rowelement ++ (["."] : (drop (x) rowelement))

我想附加列表＆＃34;取（x-1）rowelement＆＃34;和＆＃34; [＆＃34;。＆＃34;]＆＃34;和＆＃34; drop（x）rowelement＆＃34;，列表的类型将是[[a]]

Couldn't match expected type ‘Char’ with actual type ‘[Char]’
  In the expression: "."
  In the first argument of ‘(:)’, namely ‘["."]’
  In the second argument of ‘(++)’, namely
    ‘(["."] : (drop (x) rowelement))’ Failed, modules loaded: none.

x = 2 .

我知道问题是＆＃34; [＆＃34;。＆＃34;]＆＃34;但我真的不知道如何修复它。

真正的回报应该是["X","."," "," ",..,"X" ]

Answer 1

在GHCI中，您可以使用:t somefunction获取somefunction的类型。

这里问题出在(:)，让我们看看GHCI告诉我们的内容。

Prelude λ> :t (:)
(:) :: a -> [a] -> [a]

因此(:)获取a，a列表并返回新列表。专门针对手头的用例，(:)的类型为[Char] -> [[Char]] -> [[Char]]（因为a = [Char]）。但["."]的类型为[[Char]]，因此与(:)的预期不符。

现在，如果您使用take (x-1) rowelement ++ ("." : (drop (x) rowelement))（请注意[]周围缺少"."），该函数应该可以正常编译。

Answer 2

使用：

(take (x-1) rowelement) ++ ["."] ++ (drop x rowelement)

或者

(take (x-1) rowelement) ++ ("." : (drop x rowelement))

[和]不是必需的

Haskell一起列出了几个列表

2 个答案: