You can use floor in a Rational directly, no need to take it apart and divide.
There is no need to write (toRational 1), just write 1.
Don't write (subtract ai a), write (ai - i).
You also have a type error; the ai should no be a Rational, so you
need to move to toRational call to the comparison.
-- Lennart
2009/3/29 michael rice
Hi,
Thanks again for the help last night.
The second function cf2 is an attempt to reverse the process of the first function, i.e., given a rational number it returns a list of integers, possibly infinite, but you shouldn't get into trouble if you use 98%67 as input (output should be [1,2,6,5]). The interpreter is complaining about the '=' following the 'in' keyword. Is there a better way to state this?
Michael
import Data.Ratio cf :: [Int] -> Rational cf (x:[]) = toRational x cf (x:xs) = toRational x + 1 / cf xs
cf2 :: Rational -> [Int] cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) in if a = ai then [a] else ai : cf2 ((toRational 1) / (subtract ai a))
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe