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

One way to avoid explicit recursion is to use fix to express the recursion as a fixed point.

import Data.Function (fix)

fibs = fix $ \l -> [1,1] ++ zipWith (+) l (tail l)

or in point-free style

import Data.Function (fix)
import Control.Monad.Instances

fibs = fix $ ([1,1] ++) . (zipWith (+) =<< tail)