Justin Bailey jgbailey at gmail.com
Wed Oct 15 19:02:50 EDT 2008

```n decreases on each step of the recursion, which will allow it to
terminate. You need to expand AND substitute arguments:

fix (\rec n -> if n == 0 then 1 else n * rec (n-1)) 5
> fix (\rec 5 -> if 5 == 0 then 1 else n * rec (5 -1))
> fix (\rec 5 -> if 5 == 0 then 1 else n * (fix (\rec 4 -> if 4 == 0 then 1 else 4 * rec (3-1))))

And so on.

On Wed, Oct 15, 2008 at 3:51 PM, Matthew J. Williams
> hello listers, a few days ago A fellow lister sent me the following link:
>
>
>        The 'fix' function is interesting to say the least. There is one
> example that I've had difficulty expanding:
>
>        fix (\rec n -> if n == 0 then 1 else n * rec (n-1)) 5
>        120
>
>        My interpretation:
>        fix (\rec n -> if n == 0 then 1 else n * rec (n-1)) 5
>        ((\rec n -> if n == 0 then 1 else n * rec (n-1)) (fix (\rec n -> if n
> == 0 then 1 else n * rec (n-1)) )) 5
>        . . .
>
>        Yet, it does not quite explain how 'fix' does not result in infinite
> recursion.
>
>        Sincerely
>        Matthew J. Williams
>
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