Hi,
On Fri, 11 Feb 2005 12:02:56 +0100, Thomas Jäger
iii) As a side effects of how n+k patterns work, each instance of the Num class must also be an instance of Eq, which of course doesn't make sense for all numeric types.
Well this is not entirely true. I don't think 'n+k' patterns need equality, but they need ordering f (n + k) = e is like: f x | x > k = let n = x - k in e Literal patterns need equality: f 2 = e is like: f x | x == 2 = e These do not force the 'Num' class to be a superclass of 'Ord' or 'Eq'. If 'Num' was not a superclass of 'Eq', whenver you used a literal pattern there would be an extra constraint that there should be equality on the corresponding type. For example: f 2 = "Hello" would get the type: f :: (Eq a, Num a) => a -> String Hope this helps -Iavor