[Haskell] help with some basic code that doesn't work

[Matthew Walton]

pablo daniel rey wrote:
> hello
> i'm new to haskell so i'm sorry if this is a stupid question, but i'm having problems with some basic code.
> the code :
> 
> data Maybe Dir = Just Dir | Nothing
> data Dir = Left | Right | Up | Down
> data Piece = Vertical | Horizontal | CodeA | CodeB
> 
> flow = [(Horizontal, Left, Left),
>      (Horizontal, Right, Right),
>      (Vertical, Down, Down),
>      (Vertical, Up, Up), ................ etc ]
> 
> 
> fst :: (a,b,c) -> a
> fst (x,y,z) = x
> 
> scnd :: (a,b,c) -> b
> scnd (x,y,z) = y
> 
> third :: (a,b,c) -> c
> third (x,y,z) = z
> 
> element :: [(Piece, Dir, Dir)] -> Maybe Dir
> element [] = Nothing
> element xs = Just (third (head xs))
> 
> chgDir :: Piece -> Dir -> Maybe Dir
> chgDir p d = element (filter (\x -> p == (fst x)) (filter (\x -> d == (scnd x)) flow))
> 
> 
> the error i get :
> 
> Instances of (Eq Dir, Eq Piece) required for definition of chgDir

Because you're using the function (==) to compare two things of type 
Piece, Haskell needs to know how to compare them. This information isn't 
provided by your type definitions.

(==) works on types which are members of the Eq typeclass. You can 
define this instance manually, or, since your type is nice and simple, 
you can get Haskell to derive it for you.

data Piece = Vertical | Horizontal | CodeA | CodeB
              deriving Eq

similarly

data Dir = Left | Right | Up | Down
            deriving Eq

this should sort that problem out.

Incidentally, you don't need to define 'Maybe Dir'. The Maybe type is 
built in as 'Maybe a', where 'a' is a parameter which can be any type 
(such as Dir). Your functions should work unchanged if you take out your 
definition of 'Maybe Dir'.