Understanding Arrows in Haskell

Solution 1:

Please take a look in http://www.cs.yale.edu/homes/hudak/CS429F04/AFPLectureNotes.pdf, which explains how Arrows work in FRP.

2-tuples are used in defining Arrows because it's needed to represent an arrowized function taking 2 arguments.

In FRP, constants and variables are often represented as arrows which ignores its "input", e.g.

twelve, eleven :: Arrow f => f p Int
twelve = arr (const 12)
eleven = arr (const 11)

Function applications are then turned into compositions (>>>):

# (6-) 12

arr (6-) <<< twelve

Now how do we turn a 2-argument function into an arrow? For instance

(+) :: Num a => a -> a -> a

due to currying we may treat this as a function returning a function. So

arr (+) :: (Arrow f, Num a) => f a (a -> a)

now let's apply it to a constant

arr (+)             -- # f     a (a -> a)
  <<< twelve        -- # f b Int
                      :: f b     (Int -> Int)

+----------+      +-----+      +--------------+
| const 12 |----> | (+) |  ==  | const (+ 12) |
+----------+      +-----+      +--------------+

hey wait, it doesn't work. The result is still an arrow that returns a function, but we expect something akin to f Int Int. We notice that currying fails in Arrow because only composition is allowed. Therefore we must uncurry the function first

uncurry :: (a -> b -> c) -> ((a, b) -> c)

uncurry (+) :: Num a => (a, a) -> a

Then we have the arrow

(arr.uncurry) (+) :: (Num a, Arrow f) => f (a, a) a

The 2-tuple arises because of this. Then the bunch functions like &&& are needed to deal with these 2-tuples.

(&&&) :: f a b -> f a d -> f a (b, d)

then the addition can be correctly performed.

(arr.uncurry) (+)        -- # f   (a,    a) a
  <<<     twelve         -- # f b  Int
      &&& eleven         -- # f b      Int
                           :: f b           a

+--------+
|const 12|-----.
+--------+     |       +-----+      +----------+
              &&&====> | (+) |  ==  | const 23 |
+--------+     |       +-----+      +----------+
|const 11|-----'
+--------+

(Now, why don't we need things like &&&& for 3-tuples for functions having 3 arguments? Because a ((a,b),c) can be used instead.)

Edit: From John Hughes's original paper Generalising Monads to Arrows, it states the reason as

4.1 Arrows and Pairs

However, even though in case of monads the operators return and >>= are all we need to begin writing useful code, for arrows the analogous operators arr and >>> are not sufficient. Even the simple monadic addition function that we saw earlier

   add :: Monad m => m Int -> m Int -> m Int
   add x y = x >>= \u -> (y >>= \v -> return (u + v))

cannot yet be expressed in an arrow form. Making dependence on an input explicit, we see that an analogous definition should take the form

   add :: Arrow a => a b Int -> a b Int -> a b Int
   add f g = ...

where we must combine f and g in sequence. The only sequencing operator available is >>>, but f and g do not have the right types to be composed. Indeed, the add function needs to save the input of type b across the computation of f, so as to be able to supply the same input to g. Likewise the result of f must be saved across the computation of g, so that the two results can eventually be added together and returned. The arrow combinators so far introduced give us no way to save a value across another computation, and so we have no alternative but to introduce another combinator.