Re: [Haskell-cafe] [Haskell-beginners] No instance for (Show a)

31 Jul 2011

      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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