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
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