更新列表的第x个元素 - Haskell [英] Update 'x'th element of list - Haskell

问题描述

可能存在重复：

I已经在我的任务的这一部分取得了一些进展，但附加了我已经完成的部分代码：

  module网格其中
 
 data State = On |关闭导出（Eq，Show）
 
 next ::状态 - > State 
 next On = Off 
 next Off = On 
 
 type Row = [State] 
 
 
 updateRow :: Row  - > Int  - >行
 updateRow（r：rs）x 
 | x == 0 = next r：rs 
  -  |否则= ........？ 
  

如上面最后一行所示，我设法让updateRow工作，当x = 0，如下所示（倒数第0个元素）。

  * Grid> updateRow [关，关，关，关] 0 
 [开，关，关，关] 
 *网格> 
  

但是，当我尝试反转此列表中的其他元素时，我似乎无法在这个函数中'genralise'一个公式。

我也必须遵循这个类型约定：

  updateRow :: Row  - > Int  - >行
  

预先致谢。

解决方案

 模块网格其中
 
数据状态= On |关闭导出（Eq，Show）
 
 next ::状态 - > State 
 next On = Off 
 next Off = On 
 
 type Row = [State] 
 
 
 updateRow :: Row  - > Int  - >行
 updateRow（r：rs）x 
 | x == 0 = next r：rs 
 |否则= r：（updateRow rs（x-1））
 updateRow [] x = [] 
  

Possible Duplicate:
Replace individual list elements in Haskell?

I have managed to make some progress in this part of my assignment but have attached part of the code below that I have made:

module Grid where

data State = On | Off deriving (Eq, Show)

next :: State -> State
next On = Off
next Off = On

type Row = [State]


updateRow :: Row -> Int -> Row
updateRow  (r:rs) x 
    | x == 0     = next r:rs
--  | otherwise     = ........????

As shown in the last line just above, I have managed to get updateRow to work for when x = 0, as shown below (with the 0th element inverted).

*Grid> updateRow [Off,Off,Off,Off] 0
[On,Off,Off,Off]
*Grid> 

It all comes unstuck however when I try inverting other elements of this list. I can't seem to 'genralise' a formula in this function.

I also MUST follow THIS type convention:

updateRow :: Row -> Int -> Row

Thanks in advance.

解决方案

Something like that:

module Grid where

data State = On | Off deriving (Eq, Show)

next :: State -> State
next On = Off
next Off = On

type Row = [State]


updateRow :: Row -> Int -> Row
updateRow  (r:rs) x 
    | x == 0     = next r:rs
    | otherwise  = r : (updateRow rs (x-1))
updateRow [] x = []

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