[Haskell-beginners] Understanding cached fibonnacci function

Hi all, I recently stumbled on this example in some wiki: mfib :: Int -> Integer mfib = (map fib [0 ..] !!) where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) I don't understand how the results get cached. When mfib is recursively called, doesn't a new (map fib [0 ..] !!) start over again? Or perhaps I'm thinking too imperatively here... Also, if I change the definition to this (adding "a" on both sides): mfib :: Int -> Integer mfib a = (map fib [0 ..] !!) a where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) the funtion becomes slow again. Why is that? Thanks a lot, Patrick LeBoutillier -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada