Traversable instance for ((,) o) ?

[Conal Elliott]

Thanks much for the answers, Conor. They all make sense to me, particularly
about the typical information discarding in Foldable.

Regards, - Conal

On Tue, Jan 3, 2012 at 3:30 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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