CODEWITHSUNDEEP
haskellmonads
chronologos
chronologos

Reputation: 338

MonadRandom m => [g (m a)] -> m [g a]

g a is a concrete type, I am thinking of doing [g (m a)] -> [m (g a)] -> m [g a], and I know the last step can be done with sequence :: Monad m => t (m a) -> m (t a). How would I achieve the first step?

Upvotes: 1

Views: 77

Answers (1)

Daniel Wagner
Daniel Wagner

Reputation: 153162

It's pretty straightforward:

sequenceLEdge :: Functor f => LEdge (f a) -> f (LEdge a)
sequenceLEdge (l, r, act) = fmap (\v -> (l, r, v)) act

Probably there should be Foldable and Traversable instances for triples and larger tuples so you don't need to write this by hand, but there aren't for now, so...

Upvotes: 4

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