I'm having trouble embedding unconstrained monads into the NewMonad:
> {-# LANGUAGE ...,UndecidableInstances #-}
>
> instance Monad m => Suitable m v where
> data Constraints m v = NoConstraints
> constraints _ = NoConstraints
>
> instance Monad m => NewMonad m where
> newReturn = return
> newBind x k =
> let list2Constraints = constraints result
> result = case list2Constraints of
> NoConstraints -> (x >>= k)
> in result
SetMonad.hs:25:9:
Conflicting family instance declarations:
data instance Constraints Set val -- Defined at SetMonad.hs:25:9-19
data instance Constraints m v -- Defined at SetMonad.hs:47:9-19
Since Set is not an instance of Monad, there is no actual overlap
between (Monad m => m) and Set, but it seems that Haskell has no way of
knowing that.
Is there some trick (e.g. newtype boxing/unboxing) to get all the
unconstrained monads automatically instanced? Then the do notation could
be presumably remapped to the new class structure.
Dan
Wolfgang Jeltsch wrote:
Am Montag, 24. März 2008 20:47 schrieb Henning Thielemann:
[…]
Here is another approach that looks tempting, but unfortunately does not
work, and I wonder whether this can be made working.
module RestrictedMonad where
import Data.Set(Set)
import qualified Data.Set as Set
class AssociatedMonad m a where
class RestrictedMonad m where
return :: AssociatedMonad m a => a -> m a
(>>=) :: (AssociatedMonad m a, AssociatedMonad m b) =>
m a -> (a -> m b) -> m b
instance (Ord a) => AssociatedMonad Set a where
instance RestrictedMonad Set where
return = Set.singleton
x >>= f = Set.unions (map f (Set.toList x))
[…]
The problem is that while an expression of type
(AssociatedMonad Set a, AssociatedMonad Set b) =>
Set a -> (a -> Set b) -> Set b
has type
(Ord a, Ord b) => Set a -> (a -> Set b) -> Set b,
the opposite doesn’t hold.
Your AssociatedMonad class doesn’t provide you any Ord dictionary which you
need in order to use the Set functions. The instance declaration
instance (Ord a) => AssociatedMonad Set a
says how to construct an AssociatedMonad dictionary from an Ord dictionary but
not the other way round.
But it is possible to give a construction of an Ord dictionary from an
AssociatedMonad dictionary. See the attached code. It works like a
charm. :-)
Best wishes,
Wolfgang
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