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

The foldl and foldr functions are list-consumers. As svenningsson's answer rightly points out, unfoldr is a list-producer which is suitable for capturing the co-recursive structure of fibs.

However, given that foldl and foldr are polymorphic in their return types, i.e. what they're producing by consuming a list, it is reasonable to ask whether they might be used to consume one list and produce another. Might any of these produced lists be infinite?

Looking at the definition of foldl

foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f a []        = a
foldl f a (b : bs)  = foldl f (f a b) bs

we see that for foldl to produce anything at all, the list it consumes must be finite. Thus if foldl f a produces infinite output, it is because a is infinite or because f sometimes performs infinite list generation.

It's a different story with foldr

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

which admits the lazy possibility that f might generate some output for each b consumed from the input. Operations like

map g = foldr (\ b gbs -> g b : gbs) []   -- golfers prefer ((:) . g)
stutter = foldr (\ x xxs -> x : x : xxs) []

producing a little bit of output for each input, deliver infinite output from infinite input.

A cheeky person can thus express any infinitary recursion as a non-recursive foldr on an infinite list. E.g.,

foldr (\ _ fibs -> 1 : 1 : zipWith (+) fibs (tail fibs)) undefined [1..]

(Edit: or, for that matter

foldr (\_ fib a b -> a : fib b (a + b)) undefined [1..] 1 1

which is closer to the definition in the question.)

although this observation, whilst true, is hardly indicative of a healthy programming style.