Re: foldr

On Fri, Jan 02, 2004 at 02:54:11AM +0000, James Ealing wrote:

Hi all,

If I have a function:

it f [a0, a1, a2, ...] = [a0, f a1, f (f a2), ...]

Is there any way of expressing

it f

as an instance of foldr?

You can write it as it1 f l = foldr (\x xs -> x : map f xs) [] l but it's quite ineffective (the list is rewritten many times). One more effective (and quite tricky) version is it2 f l = foldr (\x g h -> h x : g (f . h)) (const []) l id But it seems more natural and effective to not use foldr at all: it3 f l = zipWith ($) (iterate (f .) id) l Best regards, Tom -- .signature: Too many levels of symbolic links