What is happening when I compose * with + in Haskell?

Question

I\'m trying to understand the result of

(*) . (+) 

in Haskell. I know that the composition operator is just the standard composition of ma

Answer 1

I understand how you feel. I found function composition to be quite difficult to grasp at first too. What helped me grok the matter were type signatures. Consider:

(*) :: Num x => x -> x -> x
(+) :: Num y => y -> y -> y
(.) :: (b -> c) -> (a -> b) -> a -> c

Now when you write (*) . (+) it is actually the same as (.) (*) (+) (i.e. (*) is the first argument to (.) and (+) is the second argument to (.)):

(.) :: (b -> c) -> (a -> b) -> a -> c
       |______|    |______|
           |           |
          (*)         (+)

Hence the type signature of (*) (i.e. Num x => x -> x -> x) unifies with b -> c:

(*) :: Num x => x -> x -> x -- remember that `x ->  x -> x`
                |    |____| -- is implicitly `x -> (x -> x)`
                |       |
                b ->    c

(.) (*) ::          (a -> b) -> a ->    c
                          |             |
                          |          |‾‾‾‾|
           Num x =>       x          x -> x

(.) (*) :: Num x => (a -> x) -> a -> x -> x

Hence the type signature of (+) (i.e. Num y => y -> y -> y) unifies with Num x => a -> x:

(+) :: Num y => y -> y -> y -- remember that `y ->  y -> y`
                |    |____| -- is implicitly `y -> (y -> y)`
                |       |
       Num x => a ->    x

(.) (*) (+) ::  Num x                => a ->     x    ->    x
                                        |        |          |
                                        |     |‾‾‾‾|     |‾‾‾‾|
                              Num y  => y     y -> y     y -> y

(.) (*) (+) :: (Num (y -> y), Num y) => y -> (y -> y) -> y -> y

I hope that clarifies where the Num (y -> y) and Num y come from. You are left with a very weird function of the type (Num (y -> y), Num y) => y -> (y -> y) -> y -> y.

What makes it so weird is that it expects both y and y -> y to be instances of Num. It's understandable that y should be an instance of Num, but how y -> y? Making y -> y an instance of Num seems illogical. That can't be correct.

However, it makes sense when you look at what function composition actually does:

( f  .  g ) = \z ->  f  ( g  z)

((*) . (+)) = \z -> (*) ((+) z)

So you have a function \z -> (*) ((+) z). Hence z must clearly be an instance of Num because (+) is applied to it. Thus the type of \z -> (*) ((+) z) is Num t => t -> ... where ... is the type of (*) ((+) z), which we will find out in a moment.

Therefore ((+) z) is of the type Num t => t -> t because it requires one more number. However, before it is applied to another number, (*) is applied to it.

Hence (*) expects ((+) z) to be an instance of Num, which is why t -> t is expected to be an instance of Num. Thus the ... is replaced by (t -> t) -> t -> t and the constraint Num (t -> t) is added, resulting in the type (Num (t -> t), Num t) => t -> (t -> t) -> t -> t.

The way you really want to combine (*) and (+) is using (.:):

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

Hence (*) .: (+) is the same as \x y -> (*) ((+) x y). Now two arguments are given to (+) ensuring that ((+) x y) is indeed just Num t => t and not Num t => t -> t.

Hence ((*) .: (+)) 2 3 5 is (*) ((+) 2 3) 5 which is (*) 5 5 which is 25, which I believe is what you want.

Note that f .: g can also be written as (f .) . g, and (.:) can also be defined as (.:) = (.) . (.). You can read more about it here:

What does (f .) . g mean in Haskell?

Hope that helps.