Composing function composition: How does (.).(.) work?

Question

(.) takes two functions that take one value and return a value:

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


        

Answer 1

Here is a simpler example of the same phenomenon:

id :: a -> a
id x = x

The type of id says that id should take one argument. And indeed, we can call it with one argument:

> id "hello" 
"hello"

But it turns out what we can also call it with two arguments:

> id not True
False

Or even:

> id id "hello"
"hello"

What is going on? The key to understanding id not True is to first look at id not. Clearly, that's allowed, because it applies id to one argument. The type of not is Bool -> Bool, so we know that the a from id's type should be Bool -> Bool, so we know that this occurrence of id has type:

id :: (Bool -> Bool) -> (Bool -> Bool)

Or, with less parentheses:

id :: (Bool -> Bool) -> Bool -> Bool

So this occurrence of id actually takes two arguments.

The same reasoning also works for id id "hello" and (.) . (.).