为什么申请工作仅针对Maybe默认

Question

I am trying to understand why do applicative functors work by default (no implementation needed) for some functors like Maybe but for others don't:

Example:
Just (+3) <*> (Just 3) works fine "out of the box"- > 6
Left (+3) <*> Left 3 does not work
Just (+3) <*> Left 4 does not work even if i declare an Either Int Int .

I assume in 99% of cases when dealing with pairs of : (f (a->b) , fa) you must implement the desired behaviour yourself ( Cartesian Product (f (a->b)) X (fa) ) and the first example is just something simple out of the box.

Example In the case of (Maybe (a->b) , Either cd) we would need to cover all 4 cases:
Just - Left Just - Right Nothing - Left Nothing -Right

Am i right in this assumption ?

Answer 1

The Applicative instance for Either is defined as:

instance Applicative (Either e) where ...

given the type of (<*>) is Applicative f => f (a -> b) -> fa -> fb for Either that is:

Either e (a -> b) -> Either e a -> Either e b

The type of Left is e -> Either ea so the type of Left (+3) is

Num a => Either (a -> a) b

and the type of Left 3 is:

Num a => Either a b

which leads to the type for Left (+3) <*> Left 3 as (Num a, Num (a -> a)) => Either (a -> a) b which is unlikely to be what you intended.

Since it's the type b which contains the function and value to operate on, using the Right constructor does work:

Right (+3) <*> Right 3
=> Right 6