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

[Ovidiu Deac]

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
<mark.spezzano at chariot.net.au> 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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>
>