> 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@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 a r | r-> a where
>
> > > build' :: [a] -> a -> r
>
> > >
>
> > > instance BuildList a [a] where
>
> > > build' l x = reverse$ x:l
>
> > >
>
> > > instance BuildList a r => BuildList a (a->r) where
>
> > > build' l x y = build'(x:l) y
>
> > >
>
> > > --build :: forall r a. (BuildList a r) => 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' l x = 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' l x = \ 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' l x = \ 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' l x = \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
>
>
>