It worked. Initially I didn't understand what you mean but after some googleing I figured it out what I had to do so I did this instance Show a ⇒ Show (Stack a) where show s = ... Now, thinking about this, it totally makes sense because Stack cannot be an instance of Show if the type a is not instance or Show. Thanks again! On Sun, Jul 31, 2011 at 12:26 PM, Mark Spezzano wrote:

Hi,

You might need a class constraint.

instance (Show a) => Show (Stack a) where

This basically lets Haskell know that if your "a" type is Showable then "Stack a" is also Showable.

Let me know if this works.

Cheers,

Mark Spezzano

On 31/07/2011, at 6:49 PM, Ovidiu Deac wrote:

...

For some reason ghc complains about not being able to call show on an Integer (?!?!?)

Please enlighten me!

ovidiu

This is the hspec:    it "shows one element"        ( show (push 1 EmptyStack) ≡ "Stack(1)")

...this is the code:

data Stack a =    EmptyStack |    StackEntry a (Stack a)    deriving(Eq)

instance Show (Stack a) where    show s =        "Stack(" ⊕ (showImpl s) ⊕ ")"        where            showImpl EmptyStack = ""            showImpl (StackEntry x _) = show x

...and this is the error:

src/Stack.hs:12:22:    No instance for (Show a)      arising from a use of `showImpl'    In the first argument of `(++)', namely `(showImpl s)'    In the second argument of `(++)', namely `(showImpl s) ++ ")"'    In the expression: "Stack(" ++ (showImpl s) ++ ")"

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