Re: [Haskell-cafe] Monad instance for partially applied type constructor?

On 29 September 2010 20:48, Ryan Ingram wrote:

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.

LiberalTypeSynonyms lets you partially apply type synonyms.