# [Haskell-beginners] Understanding cached fibonnacci function

Rafael Gustavo da Cunha Pereira Pinto RafaelGCPP.Linux at gmail.com
Fri Jan 30 04:39:08 EST 2009

```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 at 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
>
> --
> =====================
> Patrick LeBoutillier
> _______________________________________________
> Beginners mailing list
>

--
Rafael Gustavo da Cunha Pereira Pinto
Electronic Engineer, MSc.
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