RE: Counting occurrences question

===== Original Message From "xoo" ===== hi.. i was just wondering if some body could give a simple equation for the following situation.other than recursion plz..

occurrences :: Eq a => a -> [a] -> [a] --occurrences xs ys returns the number of times that xs occurs in ys

You may find it easier if you make occurrences :: Eq a => a -> [a] -> Integer since it would seem it is to return a number, and not another list! Also I would guess the function will have a form something like occurrences x xs = foldr (countOp x) 0 xs where countOp :: Eq a => a -> a -> Integer -> Integer ... I'm not sure if I should go any further, in case this is a homework question... Andy -- [ Andy Fugard ] [ +44 (0)7901 603075 ]