Re: [Haskell-cafe] (no subject)

Your Monad instance does not correspond to your Applicative instance as required by the class laws. Your Monad instance never appends the way the Applicative one does. And there's no way to make it do so. I don't think you've defined your type the way you really want. But without many details about how you wish to use it, it's hard to say how to fix the problem. On Aug 16, 2016 5:56 PM, "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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