[Haskell-cafe] Ord type on lists

[Lemmih]

On Apr 3, 2005 12:41 AM, Lloyd Smith <lloyd.g.smith at gmail.com> wrote:
> Hi everyone, I defined my own list datatype and then tried to declare
> it as an instance of type class Ord. However when I test it with
> 
> Nil > Cons 1(Nil)
> I get an "ERROR - Control stack overflow"
> 
> I am under the impression that the ord class defines default
> implementations of (<=), (>),(>=) so that I only have to supply the
> implementation of (<) shown below. Can some one tell me why this does
> not work the way I expect it to?
> 
> My very own list data type
> 
> > data List a = Nil
> >           | Cons a (List a)
> >             deriving (Show)
> 
> Derive the equality type for list.
> 
> > instance (Eq a) => Eq (List a) where
> >     Nil == Nil                          = True
> >     Nil == Cons y ys              = False
> >     Cons x xs == Nil              = False
> >     Cons x xs == Cons y ys  = (x == y) && (xs == ys)
> 
> Derive the ordered type for list
> 
> > instance (Ord a) => Ord (List a) where
> >     Nil < Nil                    = False
> >     Nil < Cons y ys       = True
> >     Cons x xs < Nil       = False
> >     Cons x xs < Cons y ys = (x < y) && (xs < ys)

(<=), (>) and (>=) are defined using 'compare', not (<). Write
'compare' yourself and everything will be good.
But why don't you just derive Eq and Ord for List?

-- 
Friendly,
  Lemmih