'f'在仿函数的fmap函数中表示什么？

Question

I'm looking at the following function:

fmap :: (a -> b) -> f a -> f b

and I want to understand what the 'f' is, as in ( fa or fb ). The article I am reading describes it as a 'box' but what's the actual correct name for it? Is it just a type variable? I think I'm confusing it and thinking it's a function application - which is correct?

Answer 1

Your intuition that it is a kind of function application is correct, but they are not regular functions. Instead, this is application of type constructors on the type level.

Specifically, Functors must have kind (type-of-type) * -> * which means they take one type argument and produce a concrete type * such as, for example, [Int] .

Examples of such type constructors include IO, Maybe, [], Either e and many others, and these specific examples all have valid Functor instances.

fmap (+1) [1,2,3]    :: [] Int -- also known as [Int]
  = [2,3,4]
fmap (+1) (Just 1)   :: Maybe Int
  = Just 2
fmap (+1) (Right 1)  :: Either e Int
  = Right 2
fmap (+1) (return 1) :: IO Int -- Uses Monad IO instance as well
  "=" 2

Answer 2

It's a type variable, representing the particular functor you're working in. For example IO is a functor, so you can specialize fmap to

fmap :: (a -> b) -> IO a -> IO b

Similarly you could specialize it to lists:

fmap :: (a -> b) -> [a] -> [b]