Bitraversable and Bifoldable are already in the bifunctors package. I needed them. ;)
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On Jan 3, 2012, at 6:30 PM, Conor McBride
On 3 Jan 2012, at 23:12, Conal Elliott wrote:
I wanted a Traversable instance for pairing, so I defined one:
instance Traversable ((,) o) where sequenceA (o,fa) = (o,) <$> fa
That looks right. Of course, we should really have a BiTraversable class of which (,) is an instance.
However, Foldable is a superclass of Traversable, so I get an error message:
Could not deduce (Foldable ((,) o)) from the context () arising from the superclasses of an instance declaration
The best I've thought of is the following:
instance Foldable ((,) o) where fold (_,m) = m
The best (upto efficiency considerations) is always
instance Foldable ((,) o) where foldMap = foldMapDefault
which amounts to what you chose.
SHE makes this a default superclass instance.
However, I don't like how it discards information.
But these folds always do discard information, discarding the shape information and accumulating over the contents. For ((,) o), seen as a functor, the first component is shape information and the second is the content.
Some questions:
* Why is Foldable a superclass of Traversable?
Because the constant-monoid Applicative makes every Traversable Foldable in a uniform way.
* Is there a good choice of a Foldable instance of ((,) o)?
Yes, the one you chose.
* Are there any other problems with the Traversable instance above (besides foldability)?
Nope. It's the Traversable instance which picks out exactly the contents that correspond to the elements abstracted by the Functor.
All the best
Conor
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