[Haskell-beginners] how do typeclasses work again?

Fri Feb 10 17:17:08 UTC 2017

Your `foo4`:
1) uses the instance `Isx isx a` to convert an `isx` into an `a`
2) then uses the instance `Show a` to convert an `a` into a String

The problem is that the compiler cannot infer the actual `a` type.

E.g., suppose you have the following instances:
data X = X
data Y = Y
instance Is (x,y) x where ...
instance Is (x,y) y where ...
instance Show X where ...
instance Show Y where ...

If you write "foo4 (X,Y)", the compiler can't decide which instance to use.

A solution: use AllowAmbiguousTypes as the compiler suggests and then 
use TypeApplications to select the "a" type:

foo4 :: forall a isx. (Is isx a, Show a) => isx -> String
foo4 = apply (\(i :: a) -> show a)

main = print (foo4 @Y (X,Y))

--
Sylvain

On 10/02/2017 17:27, Nicholls, Mark wrote:
> lovely
>
> so if I now go....
>
>> foo4 = apply (\i -> show i)
> And :t foo4 is...
>
>> foo4 :: (Is isx a, Show a) => isx -> String
> So add that as a type...and we get the same sort of awfulness...."could not deduce" bla bla
>
> But how do you make this disappear
>
>> foo4 = apply (\(i :: a) -> show i)
> Doesn’t work... "could not deduce" bla bla
>
> I'd instinctively like to go....
>
>> foo4 = apply (\(i :: ((Show a) => a)) -> show i)
> "Illegal qualified type: Show a => a"
>
> And this is really just
>
>> foo4 = apply show
> Where we end with "could not deduce"
>
> Sorry....I'm struggling.
>
>
>> -----Original Message-----
>> From: Beginners [mailto:beginners-bounces at haskell.org] On Behalf Of David
>> McBride
>> Sent: 09 February 2017 5:31 PM
>> To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level
>> topics related to Haskell
>> Subject: Re: [Haskell-beginners] how do typeclasses work again?
>>
>> foo2 :: (Is isx Integer) => isx -> String
>>
>>
>> isx -> String - That means that this function takes anything and returns a string.
>> Is isx Integer => - That just means that whatever isx is, there should be an Is isx
>> Integer instance that satisfies it.
>>
>> Putting those together this function takes anything and returns a string, so long
>> as the anything (isx) satisfies the constraint I isx Integer.
>>
>> But there's nothing in the type or code that says what type x actually is.  The
>> Integer in the constraint just constrains what isx can be.
>>
>> To fix it add the ScopedTypeVariables extension and try this:
>>
>> foo2 :: (Is isx Integer) => isx -> String
>> foo2 = apply (\(i :: Integer) -> "")
>>
>> Alternatively if you are using ghc 8, you can turn on TypeApplications and use
>> this:
>>
>> foo2 :: (Is isx Integer) => isx -> String
>> foo2 = apply @_ @Integer (\i -> "")
>>
>> On Thu, Feb 9, 2017 at 11:59 AM, Nicholls, Mark <nicholls.mark at vimn.com>
>> wrote:
>>> Sorry..I do haskell about once every 6 months for 2 hours...and then get on
>> with my life.
>>> I always forget some nuance of typeclasses.
>>>
>>> Consider some simple typeclass
>>>
>>>> class Is isx x where
>>>>    apply :: (x -> y) -> isx -> y
>>>
>>> We can make any type a member of it...mapping to itself
>>>
>>>> instance Is x x where
>>>>    apply f = f
>>> But we can also make a tuple a member of it...and pull the 1st member..
>>>
>>>> instance Is (x,y) x where
>>>>    apply f (x,y) = f x
>>> Weird and largey useless...but I'm playing.
>>>
>>> Then construct a function to operate on it
>>>
>>>> foo2 :: (Is isx Integer) => isx -> String
>>>> foo2 = apply (\i -> "")
>>> And...
>>>
>>> • Could not deduce (Is isx x0) arising from a use of ‘apply’
>>>        from the context: Is isx Integer
>>>          bound by the type signature for:
>>>                     foo2 :: Is isx Integer => isx -> String
>>>          at prop.lhs:51:3-43
>>>        The type variable ‘x0’ is ambiguous
>>>        Relevant bindings include
>>>          foo2 :: isx -> String (bound at prop.lhs:52:3)
>>>        These potential instances exist:
>>>          instance Is x x -- Defined at prop.lhs:41:12
>>>          instance Is (x, y) x -- Defined at prop.lhs:45:12
>>>      • In the expression: apply (\ i -> "")
>>>        In an equation for ‘foo2’: foo2 = apply (\ i -> "")
>>>
>>>
>>> What's it going on about?
>>> (my brain is locked in F# OO type mode)
>>>
>>> I've told it to expect a function "Integer -> String"...surely?
>>> Whats the problem.
>>>
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