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:
http://en.wikibooks.org/wiki/Haskell/Fix_and_recursion
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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