[Haskell-beginners] how do typeclasses work again?

Sat Feb 11 08:50:58 UTC 2017

ah

that works...now the challenge is for me to understand why.

I'll watch the SPJ video to get a feel for whats going on, I don't understand the difference 

forall a isx. (Is isx a, Show a) => isx -> String
and
(Is isx a, Show a) => isx -> String

but I'll look it up.

I think my understanding of type classes is naïve, I just thought it meant that secretly a dictionary was being passed.
the compiler would identify the specific dictionary from the call site
but this might be an OO minded error.

I also look up TypeApplications...thanks v much
________________________________________
From: Beginners [beginners-bounces at haskell.org] on behalf of Sylvain Henry [sylvain at haskus.fr]
Sent: 10 February 2017 17:17
To: beginners at haskell.org
Subject: Re: [Haskell-beginners] how do typeclasses work again?

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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MTV Networks International, MTV Networks UK & Ireland, Greenhouse, Nickelodeon Viacom Consumer Products, VBSi, Viacom Brand Solutions International, Be Viacom, Viacom International Media Networks and VIMN and Comedy Central are all trading names of MTV Networks Europe.  MTV Networks Europe is a partnership between MTV Networks Europe Inc. and Viacom Networks Europe Inc.  Address for service in Great Britain is 17-29 Hawley Crescent, London, NW1 8TT.