Hi, On 01/02/16 13:28, David Banas wrote:
I’m trying to do this:
data Tree a = Empty | Leaf a | Node (Tree a) a (Tree a)
[...]
instance (Monoid (f (Tree a))) => Traversable (Tree (f a)) where sequenceA Empty = pure Empty sequenceA (Leaf f) = Leaf <$> f sequenceA (Node t1 f t2) = (sequenceA t1) `mappend` (Leaf <$> f) `mappend` (sequenceA t2)
And I’m being told this:
The first argument of ‘Traversable’ should have kind ‘* -> *’, but ‘Tree (f a)’ has kind ‘*’ In the instance declaration for ‘Traversable (Tree (f a))’
And I don’t quite understand what I’m asking for that’s forbidden. Is it that I’m trying to declare that only a certain subset of /Trees/ are Traversable, and that’s not okay? It’s got to be all Trees or no Trees are Traversable?
Being Traversable (or indeed a Functor) is a property of type constructors (of kind * -> *), not of types (of kind *). In much the same way, the list type constructor [] is Traversable, but not the particular list type [Int]. The explicitly quantified type of `traverse` for a particular `Traversable t` is this: forall f a b . Applicative f => (a -> f b) -> t a -> f (t b) Notice that this involves `t a` and `t b` where `a` and `b` are polymorphic type variables, chosen by the caller of `traverse`. There's no way to constrain the particular types that might be used to instantiate those type variables. What are you really trying to do? If you'd like to write an instance for `Traversable Tree`, the haddocks for Traversable might help. :-) Or perhaps you'd like to use something like `Traversable (Compose Tree f)`? Hope this helps, Adam -- Adam Gundry, Haskell Consultant Well-Typed LLP, http://www.well-typed.com/