Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
On Wed, Sep 29, 2010 at 9:13 PM, Alexander Solla wrote:
> On 09/29/2010 02:15 PM, DavidA wrote:
>>>
>>> instance Monad (\v -> Vect k (Monomial v))
>>> >
>>
>> Yes, that is exactly what I am trying to say. And since I'm not allowed to
>> say
>> it like that, I was trying to say it using a type synonym parameterised
>> over v
>> instead.
>
> Why not:
>
> instance Monad ((->) Vect k (Monomial v))
No, what he's trying to say is
> instance Monad (Vect k . Monomial)
with some type-level composition for .
which would give these signatures:
> return :: forall a. a -> Vect k (Monomial a)
> (>>=) :: forall a b. Vect k (Monomial a) -> (a -> Vect k (Monomial b)) ->
> Vect k (Monomial b)
Notice that the "forall" variables are inside parentheses in the type;
this is what Haskell doesn't allow.
Of course you can
> newtype VectMonomial k a = VM { unVM :: Vect k (Monomial a) }
> instance Monad (VectMonomial k) where ...
But now you need to wrap/unwrap using VM/unVM.
-- ryan
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Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
David, Ryan Ingram wrote: >>> Haskell doesn't have true type functions; what you are really saying >>> is >>> >>> instance Monad (\v -> Vect k (Monomial v)) Daniel Fischer wrote: > I think there was a theoretical reason why that isn't allowed (making type > inference undecidable? I don't remember, I don't recall ...). Indeed: type inference in the presence of type-level lambdas requires higher-order unification, which is undecidable [1]. Cheers, Stefan [1] Gérard P. Huet: The Undecidability of Unification in Third Order Logic Information and Control 22(3): 257-267 (1973) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
On 09/29/2010 09:13 PM, Alexander Solla wrote: On 09/29/2010 02:15 PM, DavidA wrote: instance Monad (\v -> Vect k (Monomial v)) > Yes, that is exactly what I am trying to say. And since I'm not allowed to say it like that, I was trying to say it using a type synonym parameterised over v instead. Why not: instance Monad ((->) Vect k (Monomial v)) Sorry, I guess this is a bit off. I don't think you "really" want a monad. I think you want something like the dual to the reader monad (i.e, a comonad) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
On 09/29/2010 02:15 PM, DavidA wrote: instance Monad (\v -> Vect k (Monomial v)) > Yes, that is exactly what I am trying to say. And since I'm not allowed to say it like that, I was trying to say it using a type synonym parameterised over v instead. Why not: instance Monad ((->) Vect k (Monomial v)) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
On Wed, Sep 29, 2010 at 11:15 PM, DavidA wrote: > Ryan Ingram gmail.com> writes: > >> Haskell doesn't have true type functions; what you are really saying is >> >> instance Monad (\v -> Vect k (Monomial v)) >> > > Yes, that is exactly what I am trying to say. And since I'm not allowed to say > it like that, I was trying to say it using a type synonym parameterised over v > instead. It appears that GHC won't let you use partially applied type synonyms > as type constructors for instance declarations. Is this simply because the GHC > developers didn't think anyone would want to, or is there some theoretical > reason why it's hard, or a bad idea? The version of the lambda calculus (System Fc) GHC uses for its internal representation doesn't support lambdas at the type level. I've bumped up against this limitation myself, and don't know of any way to 'cheat' it (which makes sense, given that it's so fundamental). You can use newtypes, and recursive definitions (sometimes with overlap and/or fundeps) work for some cases, though they can get quite nasty in the harder ones. I have to assume upgrading the type system to support this would highly nontrivial, though I don't know exactly how high, nor what drawbacks there might be. > > > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Work is punishment for failing to procrastinate effectively. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Monad instance for partially applied type constructor?
It's hard. Here's a simple example: type Foo f = f Int class C (f :: (* -> *) -> *) where thingy :: f [] -> f IO -- Should this ever typecheck? I would say no; there's no way to unify f [] with [Int]. callThingy :: [Int] -> IO Int callThingy = thingy -- but what if you say this? instance C Foo where -- thingy :: Foo [] -> Foo IO -- therefore, -- thingy :: [Int] -> IO Int thingy (x:_) = return x Basically, allowing instances for type functions requires you to *un-apply* any type functions to attempt to do instance selection. -- ryan On Wed, Sep 29, 2010 at 2:15 PM, DavidA wrote: > Ryan Ingram gmail.com> writes: > >> Haskell doesn't have true type functions; what you are really saying is >> >> instance Monad (\v -> Vect k (Monomial v)) >> > > Yes, that is exactly what I am trying to say. And since I'm not allowed to say > it like that, I was trying to say it using a type synonym parameterised over v > instead. It appears that GHC won't let you use partially applied type synonyms > as type constructors for instance declarations. Is this simply because the GHC > developers didn't think anyone would want to, or is there some theoretical > reason why it's hard, or a bad idea? > > > ___ > 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] Re: Monad instance for partially applied type constructor?
On Wednesday 29 September 2010 23:15:14, DavidA wrote: > Ryan Ingram gmail.com> writes: > > Haskell doesn't have true type functions; what you are really saying > > is > > > > instance Monad (\v -> Vect k (Monomial v)) > > Yes, that is exactly what I am trying to say. And since I'm not allowed > to say it like that, I was trying to say it using a type synonym > parameterised over v instead. It appears that GHC won't let you use > partially applied type synonyms as type constructors for instance > declarations. Is this simply because the GHC developers didn't think > anyone would want to, or is there some theoretical reason why it's hard, > or a bad idea? > I think there was a theoretical reason why that isn't allowed (making type inference undecidable? I don't remember, I don't recall ...). For your problem, maybe data Vect k m b = Vect [(k, m b)] instance Monad (Vect k Monomial) where ... is an option? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Monad instance for partially applied type constructor?
Ryan Ingram gmail.com> writes: > Haskell doesn't have true type functions; what you are really saying is > > instance Monad (\v -> Vect k (Monomial v)) > Yes, that is exactly what I am trying to say. And since I'm not allowed to say it like that, I was trying to say it using a type synonym parameterised over v instead. It appears that GHC won't let you use partially applied type synonyms as type constructors for instance declarations. Is this simply because the GHC developers didn't think anyone would want to, or is there some theoretical reason why it's hard, or a bad idea? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
