On Fri, Dec 30, 2011 at 8:53 PM, Robert Weisser
newtype Set a = Set (a -> Bool) (version 1)
This one, certainly.
However, when I got to eqSet, I was stumped. I don't see how any function can see inside the set. I don't know enough Haskell to be certain, but it looks impossible to me. Is there something I am missing?
You really can't look inside the function to decide if they are equal. The only way of doing it would be enumerating all possible items that could be in a set and checking them. Something like eqSet s1 s2 = and [memSet s1 x == memSet s2 x | x <- allValues] class AllValues a where allValues :: [a] instance AllValues Bool where allValues = [True, False] instance AllValues Char where allValues = ['\0'..] instance AllValues Int where allValues = [0..] ++ [-1,-2..] and so on. Note that usually this is a really bad idea and perhaps it would be better to not implement eqSet at all. HTH, -- Felipe.