Point-free style and using $

Question

How does one combine using $ and point-free style?

A clear example is the following utility function:

times :: Int -> [a] -> [a]
times n xs = concat $ replicate n xs  

Just writing concat $ replicate produces an error, similarly you can't write concat . replicate concat . replicate either because concat expects a value and not a function.

So how would you turn the above function into point-free style?

Answer 1

You can use this combinator: (The colon hints that two arguments follow)

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

It allows you to get rid of the n :

time = concat .: replicate

Answer 2

You can easily write an almost point-free version with

times n  =  concat . replicate n

A fully point-free version can be achieved with explicit curry and uncurry:

times  =  curry $ concat . uncurry replicate

Answer 3

Get on freenode and ask lambdabot ;)

<jleedev> @pl \n xs -> concat $ replicate n xs
<lambdabot> (join .) . replicate

Answer 4

In Haskell, function composition is associative¹:

f . g . h == (f . g) . h == f . (g . h)

Any infix operator is just a good ol' function:

2 + 3 == (+) 2 3
f 2 3 = 2 `f` 3

A composition operator is just a binary function too, a higher-order one, it accepts 2 functions and returns a function:

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

Therefore any composition operator can be rewritten as such:

f . g == (.) f g
f . g . h == (f . g) . h == ((.) f g) . h == (.) ((.) f g) h
f . g . h == f . (g . h) == f . ((.) g h) == (.) f ((.) g h)

Every function in Haskell can be partially applied due to currying by default. Infix operators can be partially applied in a very concise way, using sections :

(-) == (\x y -> x - y)
(2-) == (-) 2 == (\y -> 2 - y)
(-2) == flip (-) 2 == (\x -> (-) x 2) == (\x -> x - 2)
(2-) 3 == -1
(-2) 3 == 1

As composition operator is just an ordinary binary function, you can use it in sections too:

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

Another interesting binary operator is $ , which is just function application:

f x == f $ x
f x y z == (((f x) y) z) == f x y z
f(g(h x)) == f $ g $ h $ x == f . g . h $ x == (f . g . h) x

With this knowledge, how do I transform concat $ replicate n xs into point-free style?

times n xs = concat $ replicate n xs
times n xs = concat $ (replicate n) xs
times n xs = concat $ replicate n $ xs
times n xs = concat . replicate n $ xs
times n    = concat . replicate n
times n    = (.) concat (replicate n)
times n    = (concat.) (replicate n) -- concat is 1st arg to (.)
times n    = (concat.) $ replicate n
times n    = (concat.) . replicate $ n
times      = (concat.) . replicate

¹ ^{Haskell is based on category theory . A category in category theory consists of 3 things: some objects , some morphisms , and a notion of composition of morphisms. Every morphism connects a source object with a target object , one-way. Category theory requires composition of morphisms to be associative. A category that is used in Haskell is called Hask , whose objects are types and whose morphisms are functions. A function f :: Int -> String is a morphism that connects object Int to object String . Therefore category theory requires Haskell's function compositions to be associative.}

Answer 5

By extending FUZxxl's answer, we got

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

(.::) :: (d -> e) -> (a -> b -> c -> d) -> a -> b -> c -> e
(.::) = (.).(.:)

(.:::) :: (e -> f) -> (a -> b -> c -> d -> e) -> a -> b -> c -> d -> f
(.:::) = (.).(.::)

...

Very nice.

Bonus

(.:::) :: (e -> f) -> (a -> b -> c -> d -> e) -> a -> b -> c -> d -> f
(.:::) = (.:).(.:)

Emm... so maybe we should say

(.1) = .

(.2) :: (c -> d) -> (a -> b -> c) -> a -> b -> d
(.2) = (.1).(.1)

(.3) :: (d -> e) -> (a -> b -> c -> d) -> a -> b -> c -> e
(.3) = (.1).(.2)
-- alternatively, (.3) = (.2).(.1)

(.4) :: (e -> f) -> (a -> b -> c -> d -> e) -> a -> b -> c -> d -> f
(.4) = (.1).(.3)
-- alternative 1 -- (.4) = (.2).(.2)
-- alternative 2 -- (.4) = (.3).(.1)

Even better.

We can also extend this to

fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)
fmap2 f = fmap (fmap f)

fmap4 :: (Functor f, Functor g, Functor h, functro i) 
   => (a -> b) -> f (g (h (i a))) -> f (g (h (i b)))
fmap4 f = fmap2 (fmap2 f)

which follows the same pattern.

It would be even better to have the times of applying fmap or (.) parameterized. However, those fmap or (.) s are actually different on type. So the only way to do this would be using compile time calculation, for example TemplateHaskell .

For everyday uses, I would simply suggest

Prelude> ((.).(.)) concat replicate 5 [1,2]
[1,2,1,2,1,2,1,2,1,2]
Prelude> ((.).(.).(.)) (*10) foldr (+) 3 [2,1]
60