[Haskell-beginners] lost a typeclass maybe?

[Thomas Jakway]

For what it's worth I think Bifunctors are more useful than one might think
given the lack of attention they get.

On Jun 29, 2017 2:51 PM, "Silent Leaf" <silent.leaf0 at gmail.com> wrote:

> hey it does seem to exist, so that would be
>
> foo :: (BiApplicative f) :: (i -> k i -> k i) -> f a b -> f (k a) (k b) ->
> f (k a) (k b)
> foo f fab fkakb = bipure f f <<$>> fab <<*>> fkakb
>
> pretty neat. i'm not sure the <<$>> operator exist, but the `ap` one does
> apparently.
> however i'm not sure that many people use BiApplicative ^^ But hey why not.
>
> don't pay attention to my code here, it's terribly typoed, i have no idea
> why i put the uppercase on the function Foo...
>
> 2017-06-29 20:44 GMT+02:00 Silent Leaf <silent.leaf0 at gmail.com>:
>
>> ah, obviously, the first parameter is meant to be (i -> k i -> k i).
>> mind you my opaqueBimap looks very peculiar...
>> if i isolate half of f a b:
>> Foo :: (i -> k i -> k i) -> f a -> f (k a) -> f (k a)
>> Foo f fa fas = lift f fa fas
>> so maybe i'd need a BiApplicative?
>>
>> 2017-06-29 20:38 GMT+02:00 Silent Leaf <silent.leaf0 at gmail.com>:
>>
>>> well, i sent once more my message too early by mistake.
>>> when i say invent IO a b, i don't actually mean an IO type, i meant
>>> just, any type you can't manually unbox via pattern matching or otherwise.
>>>
>>> 2017-06-29 20:36 GMT+02:00 Silent Leaf <silent.leaf0 at gmail.com>:
>>>
>>>> hi,
>>>>
>>>> i keep trying to find something that feels terribly obvious but i can't
>>>> make any link.
>>>>
>>>> say i have a function of the following type:
>>>>
>>>> foo :: (a, b) -> ([a], [b]) -> ([a], [b])
>>>> or perhaps more generally:
>>>> foo :: SomeClass f => f a b -> f [a] [b] -> f [a] [b]
>>>>
>>>> is SomeClass supposed to be BiFunctor or something else?
>>>> clearly, what i want to do is to combine the elements of the first pair
>>>> into the elements of the second, preferrably without pattern matching, that
>>>> is, merely in function of (:).
>>>>
>>>> i think the problem with bifunctor is that it seems to only allow the
>>>> application of both arguments in a separate fashion. but here the first
>>>> argument is in one block, that is (a,b).
>>>> i know, ofc we could do something like:
>>>> foo pair pairList = bimap (fst pair :) (snd pair:) pairList
>>>> or maybe use curry or whatever. but i'd like my pair to not need to be
>>>> unboxed!
>>>>
>>>> is there not a way to not have to manually call fst and snd? are both
>>>> of these functions typeclass methods by any chance? then we could write a
>>>> generalized function that could work for any f = (:) or any kind of
>>>> pair-like thingy. mind you i'm not sure to which extent it would keep the
>>>> opacity of the type constructor (,).
>>>>
>>>> especially, it's a bit like unboxing the Maybe type constructor: you
>>>> can do it manually by pattern matching, but when you have the exact same
>>>> issue but with IO, it's not possible anymore to unbox the underlying type
>>>> equally, i bet one could invent IO a b, in a way that you could not
>>>> just get a and b, but you could somehow implement
>>>> opaqueBimap :: (i -> k i) -> f a b -> f (k a) (k b) -> f (k a) (k b)
>>>> with here of course f = (,), k = [] or List, and (i -> k i) = (:)
>>>>
>>>
>>>
>>
>
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