[Haskell-beginners] Ambiguous type variable

[Ovidiu Deac]

I missed those paranthesis in your initial reply.

Indeed it worked once I put them around EmptyStack :: Stack[Char]

Thanks!

On Sun, Jul 31, 2011 at 12:23 AM, Daniel Fischer
<daniel.is.fischer at googlemail.com> wrote:
> On Saturday 30 July 2011, 22:57:04, Ovidiu Deac wrote:
>> I tried  {-# LANGUAGE ExtendedDefaultRules #-} and it didn't work. I
>> tried to put it both in Stack.hs and in TestStack.hs
>
> Aha. I don't have the impression ExtendedDefaultRules will become a much-
> used extension.
>
>>
>> Then I tried to be specific about the type in the test.
>>
>> Both this code:
>>          (pop EmptyStack :: Stack[Char]) ≡ (Nothing, EmptyStack)
>
> This needs parentheses since the '::' has very low fixity.
> Without parentheses it's parsed
>
>    ((pop EmptyStack) :: Stack [Char]) == ...
>
> so the type sugnature applies to the result of pop.
>
> I wrote
>
>   ( pop (EmptyStack :: Stack [Char]) == (...,...) )
>
> (extra spaces added to unclutter the outermost parentheses) so that the
> type signature applies only to EmptyStack.
>
> Getting the parentheses for expression type signatures right is notoriously
> tricky (unless you use parens in every case of the slightest doubt) at
> first, I should better have given the signature to Nothing,
>
>   ( pop EmptyStack == (Nothing :: Maybe [Char], EmptyStack) )
>
> that would have been harder to misread.
>
>> and this:
>>          (pop EmptyStack :: Stack[Char]) ≡ (Nothing :: Maybe[Char],
>> EmptyStack :: Stack[Char])
>
> Note that specifying the type of one subexpression is enough, pop's type
> then determines the others.
>