Haskell中a-> b Bool类型是什么意思？

Question

I know that type:

a -> Bool

is a function that takes as input something of type a and outputs a boolean.

What does type:

a -> b Bool

mean?

Answer 1

In b Bool , b stands for a parametrized type that takes one type parameter (in Haskell parlance, b is a type constructor of kind * -> * ), such as Maybe , IO or [] .

So a function of type a -> b Bool could for example take an Int and produce a Maybe Bool , IO Bool , [Bool] etc.

Answer 2

In this case b is a higher kinded type, ie a function from types to types. Sounds scary? Fear not! A higher kinded type is something like the list type. ([]) on it's own is not a type, but it still takes an argument to become one, so ([]) Int = [Int] is a type.

Kinds are like the types of types and the kind of a fully applied type is * . The kind of list on the other hand is * -> * . So it takes a type * (eg Int ) and returns a type (eg [Int] ). Other examples would be Maybe , IO or the function type (->) which has kind * -> * -> * because it still takes to types, an argument type and a result type (eg (->) Int Bool = Int -> Bool ).

The function that you have given is actually impossible to implement, just like a function of type a -> b . However a function we could implement is the following:

pureEq :: (Eq a, Applicative b) => a -> a -> b Bool
pureEq x y = pure (x == y)

Answer 3

It might be clearer if you write it out with explicit quantification and kind annotations. a -> b Bool means...

forall (a :: *) (b :: * -> *). a -> b Bool

b is therefore a type constructor taking a single type argument. Examples of single-argument type constructors abound: Maybe , [] , IO are all examples of things which you could use to instantiate b .

A function f :: a -> b Bool takes an argument of any type a ; its return value is the type constructor of your choice with its parameter filled in as Bool .

Parametricity tells us that there aren't any (non-⊥) terms with such a type.