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I've been digging into Haskell for the past months, I've run into a situation with monads that I am not quite sure how to handle.
I have one value of type a -> m a and a second of type m (a -> a) and I need to compose them such that the result of the first ends up as the input into the of the second and produces a m (a -> a) if possible. I've been stuck on this for the past day now and I'm not wrapping my head around it. I suppose that I am looking for a function like (a -> m a) -> m (a -> a) -> m (a -> a). I can provide a more concrete example if it makes more sense.
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Essentially, a >> b can be read like "do a then do b , and return the result of b ". It's similar to the more common bind operator >>= .
In Haskell, an applicative is a parametrized type that we think of as being a container for data of that type plus two methods pure and <*> . Consider a parametrized type f a . The pure method for an applicative of type f has type. pure :: a -> f a. and can be thought of as bringing values into the applicative.
As a syntactical convenience, do notation does not add anything essential, but it is often preferable for clarity and style. However, do is not needed for a single action, at all. The Haskell "Hello world" is simply: main = putStrLn "Hello world!"
Functor in Haskell is a kind of functional representation of different Types which can be mapped over. It is a high level concept of implementing polymorphism. According to Haskell developers, all the Types such as List, Map, Tree, etc. are the instance of the Haskell Functor.
You can't do this, in general. The problem is your result type: m (a -> a). This is a single monadic action which produces a function; but your first input has the form a -> m a, which (potentially) produces a different monadic action for each argument. So, for example, for the [] monad [a -> a] is list of functions with a fixed length, whereas a -> [a] can have a different length for each argument. So there's no way to 'push' the function type back into the m in general; see What is the general case of QuickCheck's promote function? for a related SO question.
If a -> m a would work for what you need, then you can turn your m (a -> a) argument into a -> m a using
\ x -> fmap ($ x) af
and use >=> (or <=<, it's not clear from your types) to compose the functions together.
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