Question

(.) :: (b -> c) -> (a -> b) -> a -> c
f . g = \x -> f(g x)

I don’t quite understand how to read the function type.

• (b -> c) is a function takes an argument b, returns c, i assume this is function f
• (a -> b) is a function takes an argument a, return b, i assume this is function g

not sure how are c in (b -> c) and a in (a -> b) are relate to a -> c

any help will be greatly appreciated, thanks!

Answer 1

Perhaps it’s better to parenthesise a bit more,

(.) :: (b -> c) -> (a -> b) -> (a -> c)

and read it:

(.) takes two arguments,

• one function (f) of type b -> c
• one function (g) of type a -> b

and returns a function (f . g) of type a -> c.

The argument type a of the resulting function is the type of the argument of (.)‘s second argument (g, which is first applied to the argument of f . g), and the result type of the composition is the result type of (.)‘s first argument (f, which is then applied to the result of the application of g to the argument).