Traversable instance for ((,) o) ?

[Edward Kmett]

Bitraversable and Bifoldable are already in the bifunctors package. I needed them. ;)

Sent from my iPad

On Jan 3, 2012, at 6:30 PM, Conor McBride <conor at strictlypositive.org> wrote:

> 
> 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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