2009/3/20 Brandon S. Allbery KF8NH
On 2009 Mar 20, at 18:01, Sean Bartell wrote:
For a type "a" to be Fractional requires there to be: (/) :: a -> a -> a You can't divide an Int by another Int and (in general) get a third Int. You would probably want something like a "Fractionable" typeclass, with (/) :: a -> a -> b which would result in a Rational, but Haskell doesn't have this.
...but there is (%) :: (Integral a) => a -> a -> Ratio a
Thanks, I knew about % but didn't remember about it when I was working on this sample :) So that being said, consider the following: Prelude Data.Ratio> let x = 5::Int Prelude Data.Ratio> :t x x :: Int Prelude Data.Ratio> :t (x%3) (x%3) :: Ratio Int Prelude Data.Ratio> let y = truncate (x%3) Prelude Data.Ratio> :t y y :: Integer Why does y now have the type Integer instead of Int?