[Haskell-cafe] type families and type signatures

[Sittampalam, Ganesh]

OK, thanks. I think I'm finally understanding :-)

Cheers,

Ganesh

-----Original Message-----
From: haskell-cafe-bounces at haskell.org [mailto:haskell-cafe-bounces at haskell.org] On Behalf Of Martin Sulzmann
Sent: 09 April 2008 07:21
To: Ganesh Sittampalam
Cc: Manuel M T Chakravarty; haskell-cafe at 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:
>
>> Sittampalam, Ganesh:
>
>>> 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?
>
>> 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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