OK, thanks. I think I'm finally understanding :-) Cheers, Ganesh -----Original Message----- From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of Martin Sulzmann Sent: 09 April 2008 07:21 To: Ganesh Sittampalam Cc: Manuel M T Chakravarty; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] type families and type signatures Manuel said earlier that the source of the problem here is foo's ambiguous type signature (I'm switching back to the original, simplified example). Type checking with ambiguous type signatures is hard because the type checker has to guess types and this guessing step may lead to too many (ambiguous) choices. But this doesn't mean that this worst case scenario always happens. Consider your example again type family Id a type instance Id Int = Int foo :: Id a -> Id a foo = id foo' :: Id a -> Id a foo' = foo The type checking problem for foo' boils down to verifying the formula forall a. exists b. Id a ~ Id b Of course for any a we can pick b=a to make the type equation statement hold. Fairly easy here but the point is that the GHC type checker doesn't do any guessing at all. The only option you have (at the moment, there's still lots of room for improving GHC's type checking process) is to provide some hints, for example mimicking System F style type application by introducing a type proxy argument in combination with lexically scoped type variables. foo :: a -> Id a -> Id a foo _ = id foo' :: Id a -> Id a foo' = foo (undefined :: a) Martin Ganesh Sittampalam wrote:

On Wed, 9 Apr 2008, Manuel M T Chakravarty wrote:

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Sittampalam, Ganesh:

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No, I meant can't it derive that equality when matching (Id a) against (Id b)? As you say, it can't derive (a ~ b) at that point, but (Id a ~ Id b) is known, surely?

No, it is not know. Why do you think it is?

Well, if the types of foo and foo' were forall a . a -> a and forall b . b -> b, I would expect the type-checker to unify a and b in the argument position and then discover that this equality made the result position unify too. So why can't the same happen but with Id a and Id b instead?

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The problem is really with foo and its signature, not with any use of foo. The function foo is (due to its type) unusable. Can't you change foo?

Here's a cut-down version of my real code. The type family Apply is very important because it allows me to write class instances for things that might be its first parameter, like Id and Comp SqlExpr Maybe, without paying the syntactic overhead of forcing client code to use Id/unId and Comp/unComp. It also squishes nested Maybes which is important to my application (since SQL doesn't have them).

castNum is the simplest example of a general problem - the whole point is to allow clients to write code that is overloaded over the first parameter to Apply using primitives like castNum. I'm not really sure how I could get away from the ambiguity problem, given that desire.

Cheers,

Ganesh

{-# LANGUAGE TypeFamilies, GADTs, UndecidableInstances, NoMonomorphismRestriction #-}

newtype Id a = Id { unId :: a } newtype Comp f g x = Comp { unComp :: f (g x) }

type family Apply (f :: * -> *) a

type instance Apply Id a = a type instance Apply (Comp f g) a = Apply f (Apply g a) type instance Apply SqlExpr a = SqlExpr a type instance Apply Maybe Int = Maybe Int type instance Apply Maybe Double = Maybe Double type instance Apply Maybe (Maybe a) = Apply Maybe a

class DoubleToInt s where castNum :: Apply s Double -> Apply s Int

instance DoubleToInt Id where castNum = round

instance DoubleToInt SqlExpr where castNum = SECastNum

data SqlExpr a where SECastNum :: SqlExpr Double -> SqlExpr Int

castNum' :: (DoubleToInt s) => Apply s Double -> Apply s Int castNum' = castNum

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