[Haskell-cafe] Monad instance for partially applied type
constructor?
Ryan Ingram
ryani.spam at gmail.com
Wed Sep 29 14:48:17 EDT 2010
On Wed, Sep 29, 2010 at 11:08 AM, DavidA <polyomino at f2s.com> wrote:
> Hi,
>
> I have the following code:
>
> {-# LANGUAGE TypeSynonymInstances #-}
>
> data Vect k b = V [(k,b)]
> -- vector space over field k with basis b
> -- for example, V [(5, E 1), (7, E 2)] would represent the vector 5 e1 + 7 e2
>
> data Monomial v = M [(v,Int)]
> -- monomials over variables v
> -- for example, M [(A,3), (B,5)] would represent the monomial a^3 b^5
>
> type Poly k v = Vect k (Monomial v)
> -- multivariate polynomials over field k and variables v
>
> instance Monad (Poly k) where
> return v = V [(1, M [(v,1)])]
> p >>= f = ... -- variable substitution
>
> So my thinking is:
> 1. The Monad type class is for one parameter type constructors (eg [], IO)
> 2. Poly is a two-parameter type constructor
> 3. So Poly k is a one-parameter type constructor
> 4. What I'm trying to express, that polynomials over field k are a monad,
> is true.
>
> However, GHC 6.12.2 complains:
>
> Type synonym `Poly' should have 2 arguments, but has been given 1
> In the instance declaration for `Monad (Poly k)'
>
> What is going wrong?
Haskell doesn't have true type functions; what you are really saying is
instance Monad (\v -> Vect k (Monomial v))
TypeSynonymInstances just lets you write stuff like this
type Foo = [Int]
instance C Foo where ...
instead of
type Foo = [Int]
instance C [Int] where ...
But it doesn't let you partially apply the type synonym.
On the other hand, if you did this:
newtype Compose f g a = O { unO :: f (g a) }
type Poly k = Compose (Vect k) Monomial
instance Monad (Poly k) where ...
would work, but now you have to wrap/unwrap Compose in the instance definition.
-- ryan
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