Dot Operator in Haskell: need more explanation

Question

I'm trying to understand what the dot operator is doing in this Haskell code:

sumEuler = sum . (map euler) . mkList

The entire source code is below.

My understanding

The dot operator is taking the two functions sum and the result of map euler and the result of mkList as the input.

But, sum isn't a function it is the argument of the function, right? So what is going on here?

Also, what is (map euler) doing?

Code

mkList :: Int -> [Int]
mkList n = [1..n-1]

euler :: Int -> Int
euler n = length (filter (relprime n) (mkList n))

sumEuler :: Int -> Int
sumEuler = sum . (map euler) . mkList

Answer 1

Put simply, . is function composition, just like in math:

f (g x) = (f . g) x

In your case, you are creating a new function, sumEuler that could also be defined like this:

sumEuler x = sum (map euler (mkList x))

The style in your example is called "point-free" style -- the arguments to the function are omitted. This makes for clearer code in many cases. (It can be hard to grok the first time you see it, but you will get used to it after a while. It is a common Haskell idiom.)

If you are still confused, it may help to relate . to something like a UNIX pipe. If f's output becomes g's input, whose output becomes h's input, you'd write that on the command-line like f < x | g | h. In Haskell, . works like the UNIX |, but "backwards" -- h . g . f $ x. I find this notation to be quite helpful when, say, processing a list. Instead of some unwieldy construction like map (\x -> x * 2 + 10) [1..10], you could just write (+10) . (*2) <$> [1..10]. (And, if you want to only apply that function to a single value; it's (+10) . (*2) $ 10. Consistent!)

The Haskell wiki has a good article with some more detail: http://www.haskell.org/haskellwiki/Pointfree

Answer 2

The . operator composes functions. For example,

a . b

Where a and b are functions is a new function that runs b on its arguments, then a on those results. Your code

sumEuler = sum . (map euler) . mkList

is exactly the same as:

sumEuler myArgument = sum (map euler (mkList myArgument))

but hopefully easier to read. The reason there are parens around map euler is because it makes it clearer that there are 3 functions being composed: sum, map euler and mkList - map euler is a single function.

Answer 3

sum is a function in the Haskell Prelude, not an argument to sumEuler. It has the type

Num a => [a] -> a

The function composition operator . has type

(b -> c) -> (a -> b) -> a -> c

So we have

           euler           ::  Int -> Int
       map                 :: (a   -> b  ) -> [a  ] -> [b  ]
      (map euler)          ::                 [Int] -> [Int]
                    mkList ::          Int -> [Int]
      (map euler) . mkList ::          Int ->          [Int]
sum                        :: Num a =>                 [a  ] -> a
sum . (map euler) . mkList ::          Int ->                   Int

Note that Int is indeed an instance of the Num typeclass.