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using foldr to do map

I am reading Hutton's book, programming in Haskell, and got stumped by a question. redefine map f using foldr. Since foldr is described as replacing (:) in a list with some provided operator it seemed it shouldn't be too hard.
foldr requires two components a what to do on empty list and the operator.

So I tried this:

> folmap :: (a -> b) -> [a] -> [b]
> folmap f = foldr (:.f) []

My reasoning is that you apply the : operator after f on every value in the list. (Though I may not understand the (.) I was thinking that if f = (a ->b) and g = (b -> c) then g.f would be g performed after f (or g.f is f followed by g).

Any advice??

By jdgallag at 2008-06-10 16:57 | LtU Forum | previous forum topic | next forum topic | other blogs | 9058 reads

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Correct

Your solution seems correct to me. You just need extra parens around the colon operator to please the compiler: folmap f = foldr ((:) . f) [] (comp.lang.haskell is probably a better place for such questions though).

By Neil Madden at Tue, 2008-06-10 17:13 | login or register to post comments

Right idea

Just your syntax is a little off. First of all, : is an infix operator, so you need to convert it to a prefix operator by wrapping it in parentheses (:).

Also, due to warts of Haskell's syntax, I strongly recommend putting a space before and and after the composition operator when used infix, even though in some cases you can get away without it. It's overloaded for qualified names (e.g. Module.function), and (:.) is a valid infix operator name. So writing (:.f) is equivalent to (\x f -> x :. f) thanks to section syntax. (Actually, (:.) is a constructor name, and not just a function, so you have to introduce it in a data definition and you can pattern match against it too.)

It doesn't appear that you are too far from understanding (.). :) There isn't much to it, and it's really easy to define in any number of ways:

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

etc etc. In Haskell, you have a lot of choices of syntax for the exact same thing. The only difference in these definitions is syntactic.

By Leon P Smith at Tue, 2008-06-10 17:36 | login or register to post comments

thanks, that did it

I had missed the parentheses around :

By jdgallag at Tue, 2008-06-10 23:28 | login or register to post comments