Hi, On Fri, 11 Feb 2005 12:02:56 +0100, Thomas Jäger <[EMAIL PROTECTED]> wrote: > > 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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
