In article <20060305233414.GA8685@soi.city.ac.uk>,
Ross Paterson
so I would also need
instance Functor ((->) a)
but where should it go? (The same instance occurs in Control.Monad.Reader, but that could be removed.)
The Prelude is to obvious place, since that's where both Functor and (->) are introduced.
There may also be a case for a class with ap (<*>) but not return (pure), which might better be called "Applicative".
Do you have both instances and clients for this interface?
Well, no, but ap gives you (>>) (I'd forgotten this earlier): ap :: Applicative' => f (a -> b) -> f a -> f b (>>) :: f a -> f b -> f b fa >> fb = ap_ (fmap const fb) fa Of course one must be careful about which argument gets "executed" first: ap_ :: Applicative' => f (a -> b) -> f a -> f b ap_ fab fa = ap (fmap (\a ab -> ab a) fa) fab ap and ap_ have the same type signature, but are different for such things as IO.
And can an associativity law even be stated?
Yes, (p >> q) >> r = p >> (q >> r)
Then I imagine you'll be similarly offended by Traversable not having Functor and Foldable as superclasses.
Yes, absolutely. It's not just more symbols, it's loss of intelligibility, since the relations between classes are not revealed. Code should be lucid. Traversable looks like my ExtractableFunctor? <http://hbase.sourceforge.net/haddock/Org.Org.Semantic.HBase.Category.Ext ractableFunctor.html> class Functor f => ExtractableFunctor f where fextract :: forall g a . FunctorApplyReturn g => f (g a) -> g (f a) ftolist :: f a -> [a] ftolist = ... -- can be derived from fextract (exercise for the reader) FunctorApplyReturn is just your Applicative (also Idiom elsewhere). -- Ashley Yakeley, Seattle WA WWED? http://www.cs.utexas.edu/users/EWD/