20 Dec
2005
20 Dec
'05
11:53 a.m.
Henning Thielemann wrote:
factors :: Integer -> Integer -> Bool factors m n | m == n = False | m < n = divides m n || factors (m+1) n
Btw. I find the recursion harder to understand than the explicit definition:
factors n = any (divides n) [2..(n-1)]
For what it's worth, I also found this function un-intuitive. What I'd expect a function called "factors" to do is exactly what yours does, and not what the one in the example does. Cheers, Daniel. -- /\/`) http://oooauthors.org /\/_/ http://opendocumentfellowship.org /\/_/ \/_/ I am not over-weight, I am under-tall. /