在数据类型构造函数的上下文中了解不同的monadic和应用绑定/组合器？

Question

I've been given the following piece of code, which I am trying to get my head around:

data MyExample e i = MyExample (CustomMonad e i)
   | forall b. MyExample e b :>>= (b -> CustomMonad e b)
   | forall b. (MyExample e (b -> a)) :<*> (MyExample e b)
   | forall b. (b -> a) :<$> (MyExample e b)

1) What do :>>= , :<*> and :<$> do differently, as opposed to

Monadic bind >>=

exampleFunction :: Int -> Maybe Int
exampleFunction el = Just (el + 100)
main = do 
  result <- exampleFunction >>= exampleFunction 21

Applicative combinators <*> and <$>

exampleFunction :: Int -> Maybe Int
exampleFunction el = Just el
main = do 
 result <- pure exampleFunction <$> (+) <*> (ExampleType 2) <*> (ExampleType 4)

2) Am I right in saying the following:

a) MyExample (CustomMonad ei) is constructing the CustomMonad type with e and i, then wrapping this in the MyExample context?

b) forall b. MyExample eb :>>= (b -> CustomMonad eb) forall b. MyExample eb :>>= (b -> CustomMonad eb) is taking a (MyExample ei) then taking b and inputting this into a function ( b -> CustomMonad eb ) that constructs a (CustomMonad eb) ?

c) forall b. (MyExample e (b -> a)) :<*> (MyExample eb) forall b. (MyExample e (b -> a)) :<*> (MyExample eb) is taking a MyExample constructed with a value e and a function (b -> a) and perform some kind of applicative combinator operation with a MyExample eb ?

d) forall b. (b -> a) :<$> (MyExample eb) forall b. (b -> a) :<$> (MyExample eb) is passing some result from (MyExample eb) to a function (b -> a)

Additionally, am I right in saying the use of forall b is ensuring that b is the same type throughout the operation?

Answer 1

:>>= doesn't actually do anything with its arguments; it simply creates a new value of type MyExample ei that you could later interpret as a call to >>= , but you could do practically anything you want with it using a different interpreter.