[Haskell-cafe] Restrictions on associated types for classes
Given class MyClass k where type AssociatedType k :: * Is there a way of requiring AssociatedType be of class Eq, say? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
{-# 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
On 17 Dec 2009, at 16:37, Stephen Lavelle wrote:
class MyClass k where
type AssociatedType k :: *
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
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
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
On 18/12/2009, at 00:37, Stephen Lavelle wrote: Given class MyClass k where type AssociatedType k :: * Is there a way of requiring AssociatedType be of class Eq, say? This works with -XFlexibleContexts: class Eq (AssociatedType k) = MyClass k where type AssociatedType k :: * Roman ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
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 :: *
?
Cheers,
Tom
--
Tom Schrijvers
Department of Computer Science
K.U. Leuven
Celestijnenlaan 200A
B-3001 Heverlee
Belgium
tel: +32 16 327544
e-mail: tom.schrijv...@cs.kuleuven.be
url: http://www.cs.kuleuven.be/~toms/
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
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
RE: [Haskell-cafe] Restrictions on associated types for classes
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-boun...@haskell.org
[mailto:haskell-cafe-boun...@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
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
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-boun...@haskell.org
[mailto:haskell-cafe-boun...@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
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
On 17 Dec 2009, at 15:31, 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. But isn't it a bit like having an instance Diff f = Diff (D f) ? Conor ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Restrictions on associated types for classes
| 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 http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
Ah yes, this is very very very helpful. Thanks : ) Miguel's example is not quite as idiomatic, but...for some reason I find it beguiling nonetheless. On Thu, Dec 17, 2009 at 2:36 PM, Roman Leshchinskiy r...@cse.unsw.edu.au wrote: On 18/12/2009, at 00:37, Stephen Lavelle wrote: Given class MyClass k where type AssociatedType k :: * Is there a way of requiring AssociatedType be of class Eq, say? This works with -XFlexibleContexts: class Eq (AssociatedType k) = MyClass k where type AssociatedType k :: * Roman ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
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 Hammondhttp://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/h/Hammond:Kevin.html, Simon L. Peyton Joneshttp://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/j/Jones:Simon_L=_Peyton.html, Philip Wadlerhttp://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/w/Wadler:Philip.html: Type Classes in Haskell. ACM Trans. Program. Lang. Syst. 18http://www.informatik.uni-trier.de/%7Eley/db/journals/toplas/toplas18.html#HallHJW96(2): 109-138 (1996) [2] Satish R. Thatte: Semantics of Type Classes Revisited. LISP and Functional Programming 1994http://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 simo...@microsoft.comwrote: | 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 http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
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.html, Simon L. Peyton Jones http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/j/Jones:Simon_L=_Peyton.html, Philip Wadler http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/w/Wadler:Philip.html: Type Classes in Haskell. ACM Trans. Program. Lang. Syst. 18 http://www.informatik.uni-trier.de/%7Eley/db/journals/toplas/toplas18.html#HallHJW96(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 simo...@microsoft.com mailto:simo...@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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Restrictions on associated types for classes
Oops, reverse that. The *instance* declaration allows for infinite types, the *class* declaration does not. Dan Weston wrote: 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.html, Simon L. Peyton Jones http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/j/Jones:Simon_L=_Peyton.html, Philip Wadler http://www.informatik.uni-trier.de/%7Eley/db/indices/a-tree/w/Wadler:Philip.html: Type Classes in Haskell. ACM Trans. Program. Lang. Syst. 18 http://www.informatik.uni-trier.de/%7Eley/db/journals/toplas/toplas18.html#HallHJW96(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 simo...@microsoft.com mailto:simo...@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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
