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

You're right, (.) only takes two arguments. You just seem to be confused with the syntax of haskell. In the expression (.).(.), it's in fact the dot in the middle that takes the other two dots as argument, just like in the expression 100 + 200, which can be written as (+) 100 200.

(.).(.) === (number the dots)
(1.)2.(3.) === (rewrite using just syntax rules)
(2.)(1.)(3.) === (unnumber and put spaces)
(.) (.) (.) ===

And it should be even more clear from (.) (.) (.) that the first (.) is taking the second (.) and third (.) as it's arguments.