Along the same lines: {-# LANGUAGE GADTs, TypeFamilies #-} module Assoc where data EqD k where EqD :: Eq k => EqD k class MyClass k where type AssociatedType k :: * evidence :: k -> EqD (AssociatedType k) instance MyClass () where type AssociatedType () = Integer evidence _ = EqD eq :: MyClass k => k -> AssociatedType k -> AssociatedType k -> Bool -- eq k k1 k2 = k1 == k2 -- doesn't work eq k k1 k2 = case evidence k of EqD -> k1 == k2 -- works fine On 17 Dec 2009, at 17:10, Miguel Mitrofanov wrote:

{-# LANGUAGE GADTs, TypeFamilies #-} module Assoc where data EqD k where EqD :: Eq k => EqD k class MyClass k where data AssociatedType k :: * evidence :: AssociatedType k -> EqD (AssociatedType k) eq :: MyClass k => AssociatedType k -> AssociatedType k -> Bool -- eq k1 k2 = k1 == k2 -- doesn't work eq k1 k2 = case evidence k1 of EqD -> k1 == k2 -- works fine

Hi all On 17 Dec 2009, at 14:22, Tom Schrijvers wrote:

...

class MyClass k where type AssociatedType k :: *

Is there a way of requiring AssociatedType be of class Eq, say?

Have you tried:

{-# LANGUAGE TypeFamilies #-} {-# LANGUAGE FlexibleContexts #-}

class Eq (AssociatedType k) => MyClass k where type AssociatedType k :: *

I just got very excited about this. I'm supposed to be setting a test, but this is far more interesting. I tried this

{-# LANGUAGE TypeFamilies, FlexibleContexts, EmptyDataDecls, TypeOperators #-}

module DDD where

class (Diff (D f)) => Diff f where type D f plug :: D f x -> x -> f x

newtype K a x = K a deriving Show

data Void magic :: Void -> a magic x = x `seq` error "haha"

instance Diff (K a) where type D (K a) = K Void plug (K c) x = magic c

newtype I x = I x deriving Show

instance Diff I where type D I = K () plug (K ()) x = I x

data (f :+: g) x = L (f x) | R (g x) deriving Show

instance (Diff f, Diff g) => Diff (f :+: g) where type D (f :+: g) = D f :+: D g plug (L f') x = L (plug f' x) plug (R g') x = R (plug g' x)

data (f :*: g) x = f x :& g x deriving Show

instance (Diff f, Diff g) => Diff (f :*: g) where type D (f :*: g) = (D f :*: g) :+: (f :*: D g) plug (L (f' :& g)) x = plug f' x :& g plug (R (f :& g')) x = f :& plug g' x

But I got this message [1 of 1] Compiling DDD ( DDD.lhs, interpreted ) DDD.lhs:5:2: Cycle in class declarations (via superclasses): DDD.lhs:(5,2)-(7,28): class (Diff (D f)) => Diff f where { type family D f; } Failed, modules loaded: none. and now I have to go back to setting my class test. Sorry for spam Conor

Which is 'right' in a way, since in the language Conor defined, all definable terms are "infinitely differentiable", and that can be inferred from the given rules. That, in practice, you only need a finite number of them in any given computation is derivable from the instances, but is not a general theorem. Jacques PS: wait until someone wants to define resurgent functions, where you need ordinals to keep track of similar information! Nice thing is that it's all still computable, you just have to work a fair bit harder to set up the correct machinery. See papers by Salvy & Shackell, as well as van der Hoeven. Simon Peyton-Jones wrote:

Hmm. If you have class (Diff (D f)) => Diff f where

then if I have f :: Diff f => ... f = e then the constraints available for discharging constraints arising from e are Diff f Diff (D f) Diff (D (D f)) Diff (D (D (D f))) ...

That's a lot of constraints.

Simon

| -----Original Message----- | From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On | Behalf Of Conor McBride | Sent: 17 December 2009 14:48 | To: Haskell Cafe | Subject: Re: [Haskell-cafe] Restrictions on associated types for classes | | Hi all | | On 17 Dec 2009, at 14:22, Tom Schrijvers wrote: | | >> class MyClass k where | >> type AssociatedType k :: * | >> | >> Is there a way of requiring AssociatedType be of class Eq, say? | > | > Have you tried: | > | > {-# LANGUAGE TypeFamilies #-} | > {-# LANGUAGE FlexibleContexts #-} | > | > class Eq (AssociatedType k) => MyClass k where | > type AssociatedType k :: * | | I just got very excited about this. I'm supposed to be | setting a test, but this is far more interesting. I tried | this | | > {-# LANGUAGE TypeFamilies, FlexibleContexts, EmptyDataDecls, | TypeOperators #-} | | > module DDD where | | > class (Diff (D f)) => Diff f where | > type D f | > plug :: D f x -> x -> f x | | > newtype K a x = K a deriving Show | | > data Void | > magic :: Void -> a | > magic x = x `seq` error "haha" | | > instance Diff (K a) where | > type D (K a) = K Void | > plug (K c) x = magic c | | > newtype I x = I x deriving Show | | > instance Diff I where | > type D I = K () | > plug (K ()) x = I x | | > data (f :+: g) x = L (f x) | R (g x) deriving Show | | > instance (Diff f, Diff g) => Diff (f :+: g) where | > type D (f :+: g) = D f :+: D g | > plug (L f') x = L (plug f' x) | > plug (R g') x = R (plug g' x) | | > data (f :*: g) x = f x :& g x deriving Show | | > instance (Diff f, Diff g) => Diff (f :*: g) where | > type D (f :*: g) = (D f :*: g) :+: (f :*: D g) | > plug (L (f' :& g)) x = plug f' x :& g | > plug (R (f :& g')) x = f :& plug g' x | | But I got this message | | [1 of 1] Compiling DDD ( DDD.lhs, interpreted ) | | DDD.lhs:5:2: | Cycle in class declarations (via superclasses): | DDD.lhs:(5,2)-(7,28): class (Diff (D f)) => Diff f where { | type family D f; } | Failed, modules loaded: none. | | and now I have to go back to setting my class test. | | Sorry for spam | | Conor | | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe@haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe

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| > Hmm. If you have | > class (Diff (D f)) => Diff f where | > | > then if I have | > f :: Diff f => ... | > f = e | > then the constraints available for discharging constraints arising | > from e are | > Diff f | > Diff (D f) | > Diff (D (D f)) | > Diff (D (D (D f))) | > ... | > | > That's a lot of constraints. | | But isn't it a bit like having an instance | | Diff f => Diff (D f) A little bit. And indeed, could you not provide such instances? That is, every time you write an equation for D, such as type D (K a) = K Void make sure that Diff (K Void) also holds. The way you it, when you call f :: Diff f => <blah>, you are obliged to pass runtime evidence that (Diff f) holds. And that runtime evidence includes as a sub-component runtime evidence that (Diff (D f)) holds. If you like the, the evidence for Diff f looks like this: data Diff f = MkDiff (Diff (D f)) (D f x -> x -> f x) So you are going to have to build an infinite data structure. You can do that fine in Haskell, but type inference looks jolly hard. For example, suppose we are seeking evidence for Diff (K ()) We might get such evidence from either a) using the instance decl instance Diff (K a) where ... or b) using the fact that (D I) ~ K (), we need Diff I, so we could use the instance instance Diff I Having two ways to get the evidence seems quite dodgy to me, even apart from the fact that I have no clue how to do type inference for it. Simon

I think the denotational meanings are different. The instance also implies: For each Cl t there must exist a Cl u where u does not unify with [v] for some v. In other words, there must be a ground instance. For the class declaration, the existence of a ground instance can be inferred only by excluding infinite types with strict type unification semantics. If infinite types were admitted (where type unification is done non-strictly), the class declaration allows for infinite types (let t ~ [t] in t). The instance does not. Dan Martin Sulzmann wrote:

The statements

class Cl [a] => Cl a

and

instance Cl a => Cl [a]

(I omit the type family constructor in the head for simplicyt)

state the same (logical) property:

For each Cl t there must exist Cl [t].

Their operational meaning is different under the dictionary-passing translation [1]. The instance declaration says we build dictionary Cl [a] given the dictionary Cl [a]. The super class declaration says that the dictionary for Cl [a] must be derivable (extractable) from Cl a's dictionary. So, here we run into a cycle (on the level of terms as well as type inference).

However, if we'd adopt a type-passing translation [2] (similar to dynamic method lookup in oo languages) then there isn't necessarily a cycle (for terms and type inference). Of course, we still have to verify the 'cyclic' property which smells like we run into non-termination if we assume some inductive reason (but we might be fine applying co-induction).

-Martin

[1] Cordelia V. Hall, Kevin Hammond http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/h/Hammond:Kevin...., Simon L. Peyton Jones http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/j/Jones:Simon_L=..., Philip Wadler http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/w/Wadler:Philip....: Type Classes in Haskell. ACM Trans. Program. Lang. Syst. 18 http://www.informatik.uni-trier.de/%7Eley/db/journals/toplas/toplas18.html#H...(2): 109-138 (1996)

[2] Satish R. Thatte: Semantics of Type Classes Revisited. LISP and Functional Programming 1994 http://www.informatik.uni-trier.de/%7Eley/db/conf/lfp/lfp1994.html#Thatte94: 208-219

On Thu, Dec 17, 2009 at 6:40 PM, Simon Peyton-Jones mailto:simonpj@microsoft.com> wrote:

| > Hmm. If you have | > class (Diff (D f)) => Diff f where | > | > then if I have | > f :: Diff f => ... | > f = e | > then the constraints available for discharging constraints arising | > from e are | > Diff f | > Diff (D f) | > Diff (D (D f)) | > Diff (D (D (D f))) | > ... | > | > That's a lot of constraints. | | But isn't it a bit like having an instance | | Diff f => Diff (D f)

A little bit. And indeed, could you not provide such instances? That is, every time you write an equation for D, such as type D (K a) = K Void make sure that Diff (K Void) also holds.

The way you it, when you call f :: Diff f => <blah>, you are obliged to pass runtime evidence that (Diff f) holds. And that runtime evidence includes as a sub-component runtime evidence that (Diff (D f)) holds. If you like the, the evidence for Diff f looks like this: data Diff f = MkDiff (Diff (D f)) (D f x -> x -> f x) So you are going to have to build an infinite data structure. You can do that fine in Haskell, but type inference looks jolly hard.

For example, suppose we are seeking evidence for Diff (K ()) We might get such evidence from either a) using the instance decl instance Diff (K a) where ... or b) using the fact that (D I) ~ K (), we need Diff I, so we could use the instance instance Diff I

Having two ways to get the evidence seems quite dodgy to me, even apart from the fact that I have no clue how to do type inference for it.

Simon _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org mailto:Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe