Re: [Haskell-cafe] (no subject)

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" 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 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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