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
<[email protected]> 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) ++ ")"
>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://www.haskell.org/mailman/listinfo/beginners
>
>
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