fmap (<*>) :: m (n (a -> b)) -> m (n a -> n b)
so
f <**> x = (fmap (<*>) f) <*> x
On Mon, Oct 12, 2009 at 9:22 AM, Kim-Ee Yeoh
Does anyone know if it's possible to write the following:
<**> :: (Applicative m, Applicative n) => m (n (a->b)) -> m (n a) -> m (n b)
Clearly, if m and n were monads, it would be trivial.
Rereading the original paper, I didn't see much discussion about such nested app. functors.
Any help appreciated.
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