[Attempting to resend] Hi - I think your logic is: I can define these two: fFrac :: Fractional a => a -> a fFrac n = n * (n+1) / 2 fInt :: Integral a => a -> a fInt n = n * (n+1) `div` 2 so that, e.g. fFrac (5.0 :: Float) == 15.0 :: Float fInt (5 :: Integer) == 15 :: Integer And all number types are either Integral or Fractional, so surely I should be able to define a single function of type: f :: Num a => a -> a This would seem reasonable, but I think there’s a problem with the last assumption. It is indeed possible for other types to be instances of Num, but not of Integral or Fractional. For example, I could define: instance Num Bool where fromInteger 0 = False fromInteger _ = True (+) = (&&) (*) = (||) abs = id signum _ = True negate = not Now this would probably be pretty dumb (and probably doesn’t comply with expectations<https://hackage.haskell.org/package/base-4.14.0.0/docs/Prelude.html#t:Num>), but is possible. (And also pretty dumb to define it without also define an instance Integral Bool where ..., but still possible). So I don’t think f :: Num a => a -> a could be possible, since Num by itself (& the dumb Bool instance) has no way to do the division. (At least that I can think of, but would be very interested to hear if there is). Regards, David. From: Beginners <beginners-bounces@haskell.org> on behalf of M Douglas McIlroy <m.douglas.mcilroy@dartmouth.edu> Sent: Friday, December 11, 2020 1:03:03 PM To: beginners@haskell.org <beginners@haskell.org> Subject: [Haskell-beginners] unwanted Fractional constraint For rational functions that take on integer values at integer arguments, for example n*(n+1)/2, is there a way to doctor the corresponding Haskell definition f n = n*(n+1)/2 so that the type signature becomes f :: Num a => a -> a rather than f :: Fractional a => a -> a Doug McIlroy _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
