Question

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.

Answer 1

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 = []