7 Mar
2017
7 Mar
'17
12:11 p.m.
On Tue, 7 Mar 2017, Mario Blažević wrote:
On 2017-03-07 11:40 AM, Henning Thielemann wrote:
I think it must be Monoid (m a). But it won't work, because NonEmpty is no Monoid because there is no mempty.
You're correct on both counts, sorry about that. It would have to be
genericMany, genericSome :: (Alternative f, Applicative m, Semigroup (m a)) => f a -> f (m a)
once Semigroup is in base.
But then, for genericMany, Monoid would be appropriate, again. :-)