This isn't necessarily correct, is it? An equality-based (n^2) nub
works "fine" on infinite lists, whereas any O(n log n) sort-based nub
must necessarily evaluate the entire list before being able to return
the value. The original n^2 nub also returns the elements in the order
of their first appearance rather than sorted order.
Of course, you may not care about this and just be experimenting with
rewrite rules, in which case I can't help you :)
Dan
On Wed, Jun 3, 2009 at 5:58 AM, Milan Straka
Hi,
I am interesting in writing a method nub in such a way, that it will use O(N^2) trivial algorithm for types with equality only, and O(N log N) algorithm for comparable types. I would like to say
class Nub where nub :: [a] -> [a] instance Ord a => Nub a where nub = nubOrd instance Eq a => Nub a where nub = nubEq
which is of course not valid Haskell. I tried using GHC rewriting rules to achieve this. My first try was
{-# NOINLINE nub #-} nub :: Eq a => [a] -> [a] nub xs = ...
nubOrd :: Ord a => [a] -> [a] nubOrd xs = ...
{-# RULES "nub/nubOrd" nub = nubOrd #-}
which does not work either, because of missing an Ord a context. I was not able to write the rule which would fire.
Is there a way to write such a rewriting rule or there is no way of acquiring the Ord dictionary in rewrite rule? Or does anyone know any other way of implementing such a nub without explicitly listing all Ord instances?
I wish you a nice day, Milan Straka
PS: Listing {-# RULES "nub/nubOrd Int" nub = nubOrd :: [Int]->[Int] #-} for all Ord instances would work, but ... _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users