I'm naively capable of messing around with type families...so I know how to define the types statically...that's not really what I want...that's too strong. I think I'm trying to work in the universe of typeclasses and not data types....in my OO head these two things overlap (if you see what I mean)....in Haskell they are distinct...which I'm beginning to feel makes type inference easy....but is actually quite "weak". so lets start again...
class Is isx x where apply ::(x -> y) -> isx -> y
instance Is x x where apply f = f
i.e. lets create our tuple instance like this!
instance (Is m x) => Is (m,y) x where apply f (m,y) = apply f m
data X = X deriving (Show) data Y = Y deriving (Show)
foo4 :: forall a isx. (Is isx a, Show a) => isx -> String foo4 = apply (\(i :: a) -> show i)
now...this line said....
main = print (foo4 @X (X,Y))
and that works!....which I think is what I want....in an OO world this feels like a "cast"....where Ive said (X,Y) <: X....I'm getting the compiler to extract fst for me...I'm lazy. so lets tell the compiler it could do snd for me.
instance (Is m x) => Is (y,m) x where apply f (y,m) = apply f m
gives....."Duplicate instance declarations " which is unfortunate as I wanted to then write
main = print (foo4 @Y (X,Y))
I don't understand the difference
forall a isx. (Is isx a, Show a) => isx -> String and (Is isx a, Show a) => isx -> String I have used the `forall` explicitly only to fix the order of the type
to "cast" to Y....which feels perfectly reasonable then I look this up on the interweb...and magically found some noob has been here before! "noob “Duplicate instance declarations” (again)" that noob was me!...about a year ago...and someone said I was misunderstanding how to define these sort of recursive structures....and It should be done in the class declaration...which doesn't seem to work in this case...as I want to do something recursive the answer said..... "Haskell requires that there be only one instance for each class and type. Thus is determined only from the part to the right of the =>. " ok, I buy that...IF I want to guarantee Haskell to automatically derive type class dictionaries that functional restriction is perfectly reasonable...as long as its resolved at some point. I tbink (naively)...Haskell is saying... "if I match things against (x,y) I've got 2 instance declarations....so how do I decide which one?" what I'm saying is....that's fine but I'm telling you which one at the call site....using these "@" things... so what's the problem?...the functional restriction is too restrictive. ________________________________________ From: Beginners [beginners-bounces@haskell.org] on behalf of Sylvain Henry [sylvain@haskus.fr] Sent: 11 February 2017 13:36 To: beginners@haskell.org Subject: Re: [Haskell-beginners] how do typeclasses work again? On 11/02/2017 09:50, Nicholls, Mark wrote: parameters (`a` and `isx`) so that we are sure to set the type of `a` when we write (using TypeApplications): foo4 @Y (X,Y) In the second declaration, the `forall` is implicit.
I think my understanding of type classes is naïve, I just thought it meant that secretly a dictionary was being passed. Yes your understanding is correct. The issue here is that the compiler doesn't know the type of `a`, hence it can't select and pass the appropriate instances.
the compiler would identify the specific dictionary from the call site Even at call site, the compiler can't infer the `a` type from the `isx` type (nor from the return type of `foo4`).
Do you want the `a` type to be dependent on the `isx` type? I.e., to only be allowed to define a single `Is isx a` instance for each `isx` type. -- Sylvain _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners CONFIDENTIALITY NOTICE This e-mail (and any attached files) is confidential and protected by copyright (and other intellectual property rights). If you are not the intended recipient please e-mail the sender and then delete the email and any attached files immediately. Any further use or dissemination is prohibited. While MTV Networks Europe has taken steps to ensure that this email and any attachments are virus free, it is your responsibility to ensure that this message and any attachments are virus free and do not affect your systems / data. Communicating by email is not 100% secure and carries risks such as delay, data corruption, non-delivery, wrongful interception and unauthorised amendment. If you communicate with us by e-mail, you acknowledge and assume these risks, and you agree to take appropriate measures to minimise these risks when e-mailing us. 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.