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You can reach the person managing the list at beginners-ow...@haskell.org When replying, please edit your Subject line so it is more specific than "Re: Contents of Beginners digest..." Today's Topics: 1. Re: how do typeclasses work again? (Nicholls, Mark) ---------------------------------------------------------------------- Message: 1 Date: Sat, 11 Feb 2017 08:50:58 +0000 From: "Nicholls, Mark" <nicholls.m...@vimn.com> To: "The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell" <beginners@haskell.org> Subject: Re: [Haskell-beginners] how do typeclasses work again? Message-ID: <e7e7fdf32472ff48bb0e4d9dc4283d0e8bd03...@mtvne-exmb02.mtvne.ad.viacom.com> Content-Type: text/plain; charset="windows-1252" 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-boun...@haskell.org] on behalf of Sylvain Henry [sylv...@haskus.fr] Sent: 10 February 2017 17:17 To: beginners@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-boun...@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.m...@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. >>> >>> CONFIDENTIALITY NOTICE >>> >>> This e-mail (and any attached files) is confidential and protected by >>> copyright >> (and other intellectual property rights). 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