I don't know exactly how to explain it, but it has to do with memoization The first version generates a fully memoized function, while the second creates a memoized version for each call, since the thunk and memoization for fib is destroyed after each computation. This is not a precise explanation, but I can't think of a better way to put it... On Thu, Jan 29, 2009 at 16:18, Patrick LeBoutillier < patrick.leboutillier@gmail.com> wrote:

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

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