[Haskell-beginners] From Functional Dependencies to Type Families

[builes.adolfo at googlemail.com]

> You need to tell the compiler explicitly that a and Build r should be the  
> same type.

Thanks Daniel :). That was the trick .

It's the first time that I see "~", is that from -XUndecidableInstances ? .

Also, thanks to Stephen.

-
Adolfo

On Dec 3, 2009 6:30am, Daniel Fischer <daniel.is.fischer at web.de> wrote:
> Am Donnerstag 03 Dezember 2009 04:06:43 schrieb Adolfo Builes:

> > > {-# OPTIONS -fglasgow-exts #-}

> > >

> > > module VarArg where

> > > import Data.FiniteMap -- for an example below

> > >

> > > class BuildList ar | r-> a where

> > > build' :: [a] -> a -> r

> > >

> > > instance BuildList a [a] where

> > > build' lx = reverse$ x:l

> > >

> > > instance BuildList ar => BuildList a (a->r) where

> > > build' lxy = build'(x:l) y

> > >

> > > --build :: forall r a. (BuildList ar) => a -> r

> > > build x = build' [] x

> >

> > I'm trying to replace the code below to work with type families, I  
> started

> > out replacing the definition of class with :

> >

> > class BuildList r where

> > type Build r

> > build' :: [Build r] -> Build r -> r

> >

> > follow by the instance for [a] resulting in

> >

> > instance BuildList [a] where

> > type Build [a] = a

> > build' lx = reverse $ x:l

> >

> > Until here, everything is working, and I'm able to do

> >

> > > build' [2,3,4] 1 :: [Integer]

> > > [4,3,2,1]

> >

> > then I move on to the next instance (a -> r) with

> >

> > instance BuildList r => BuildList (a-> r) where

> > type Build (a -> r) = a

> > build' lx = \ y -> build'(x:l) y

> >

> >

> > But I get the following error

> >

> > Couldn't match expected type `Build r' against inferred type `a'

> > `a' is a rigid type variable bound by

> > the instance declaration at /home/adolfo/foo.hs:347:35

> > In the first argument of `(:)', namely `x'

> > In the first argument of `build'', namely `(x : l)'

> > In the expression: build' (x : l) y

> >

> > then I try with :

> >

> > instance BuildList r => BuildList (a-> r) where

> > type Build (a -> r) = Build r

> > build' lx = \ y -> build'(x:l) y

> >

> >

> > And I get

> >

> > Couldn't match expected type `a' against inferred type `Build r'

> > `a' is a rigid type variable bound by

> > the instance declaration at /home/adolfo/foo.hs:347:35

> > Expected type: [Build r]

> > Inferred type: [Build (a -> r)]

> > In the second argument of `(:)', namely `l'

> > In the first argument of `build'', namely `(x : l)'

> >

> > I have been trying to figure out, which type should it be, but I haven't

> > found the correct one, any ideas ?



> I think



> instance (BuildList r, Build r ~ a) => BuildList (a -> r) where

> type Build (a -> r) = a

> build' lx = \y -> build' (x:l) y



> should work.



> You need to tell the compiler explicitly that a and Build r should be the  
> same type.



> >

> >

> > Thanks

> >

> > -

> > Adolfo Builes



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