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parsonsmatt avatar parsonsmatt commented on July 18, 2024 1

You are correct that the second instance is the ZipList Applicative. When a type has both a Monad and an Applicative instance, it's necessary for them to agree. That is, given ap,

ap :: Monad m => m (a -> b) -> m a -> m b
ap mf ma = do
  f <- mf
  a <- ma
  pure (f a)

then for any values, f `ap` a == f <*> a. Since ZipList does not have a Monad, we give the "cross-product-y" one to List itself.

Adding a comment to indicate two possible instances is a good idea :) Thanks

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