[Haskell-cafe] (no subject)

[Chris Wong]

Hi Corentin,

That's called the "Validation applicative". I'm on my phone so I can't
link, but that phrase should give you a starting point.

Note that your type actually has no valid Monad instance that is consistent
with the Applicative. That's because while the Applicative accumulates
data, the Monad throws it away every time.

Chris

On Aug 17, 2016 09:56, "Corentin Dupont" <corentin.dupont at gmail.com> wrote:

> That, right, just to clarify, my instances are like that:
>
> instance Applicative (Todo a) where
>    pure                = Done
>    Todo as <*> Todo bs = Todo $ as ++ bs
>    Todo as <*> _       = Todo as
>    Done f  <*> r       = fmap f r
>
> instance Monad (Todo a) where
>    return        = Done
>    Todo as >>= _ = Todo as
>    Done a >>= f  = f a
>
>
> It's basically accumulating on the Todos. Either [a] b does not fit
> because it does not accumulate on the a's.
>
>
> On Tue, Aug 16, 2016 at 11:12 PM, David Kraeutmann <kane at kane.cx> wrote:
>
>> I went from the assumption that he just wants something like
>>
>> instance Monad (Todo a)
>>
>> and that implies that b is the type inside of the monad.
>>
>>
>> On 08/16/2016 08:23 PM, David Kraeutmann wrote:
>> > This looks functionally identical to type Todo a b = Either [a] b, and
>> Either has a Monad instance.
>> > On 8/16/2016 8:21 PM, Corentin Dupont wrote:
>> >> Hi guys,
>> >> is there some library with a Monad (+Applicative, Functor...) instance
>> of
>> >> the following type:
>> >>
>> >> data Todo a b = Todo [a] | Done b
>> >>
>> >> Thanks!
>> >> Corentin
>> >>
>> >>
>> >>
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>> >
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