Can fold be used to create infinite lists?

Question

I have written the following code which creates an infinite list of Fibonacci numbers:

fibs = 1:1:fib 1 1
  where fib a b = a+b:fib b (a+b)

Answer 1

You can use zipWith to write your definition

fibonacci = 1:1:zipWith (+) fibonacci (tail fibonacci)

edit: Ok, I dont think you can use foldl or foldr to create infinite list. Not in any simple imaginable sense. If you look at the simple definition of foldl

foldl f z []     = z
foldl f z (x:xs) = foldl f (f z x) xs

foldl never returns until it has exhausted the whole list. So a simple example like

g = foldl f [] [1..]
 where 
  f xs a = xs ++ [a]

> take 10 g

will not work even and it will loop on forever.