Or you could use curry and uncurry:
sumProducts = curry $ sum . uncurry (zipWith (*))
On 5 June 2011 17:40, Daniel Fischer
On Sonntag, 5. Juni 2011, 17:09, Alexander Shendi wrote:
Hi folks,
I am working my way through "Learn You a Haskell for Great Good" and have reached page 84, where the book recommends that you define your functions in a "points-free style", using the "." and "$" operators.
Now I have:
sumProducts' :: Num a => [a] -> [a] -> a sumProducts' x y = sum (zipWith (*) x y)
I would like to eliminate the "x" and the "y" in the definition, but all I have managed to contrive is:
sumProducts :: Num a => [a] -> [a] -> a sumProducts x = sum . zipWith (*) x
How do I proceed from here? Any advice is welcome :)
Many thanks in advance,
/Alexander
First, (mentally) insert parentheses:
sumProducts x = sum . (zipWith (*) x)
Now, write the composition as a prefix application of (.),
sumProducts x = (.) sum (zipWith (*) x) = ((.) sum) (zipWith (*) x)
which has the form f (g x), with f = (.) sum and g = zipWith (*), so it's
(f . g) x, which expands to
(((.) sum) . (zipWith (*))) x
and now the argument can easily be dropped, giving
sumProducts = ((.) sum) . (zipWith (*))
now to make it look a little nicer, we can remove unnecessary parentheses and write (.) sum as a section,
sumProducts = (sum .) . zipWith (*)
Generally, pointfreeing goes
f (g x) ~> f . g f (g x y) ~> (f .) . g f (g x y z) ~> ((f .) .) . g
Play with it and get a little experience, but don't overuse it. Nobody really wants to come across a pointfree version of
\a b c d e f g h -> foo (bar a b) (baz c d e) f (quux g h)
(and that's not even repeating or flipping arguments)
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