Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

Indeed the second snipper contains quite an obvious mistake. Thanks for noticing! It doesn't seem to me it utilises a lambda expression though? You mean the '.' operator for chaining function? If that's it, it could be rewritten unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique (filter (/= x) xs) | otherwise = x : unique xs ----- Original Message ----- From: Ramesh Kumar Sent: 03/28/12 10:14 AM To: franco00@gmx.com, beginners@haskell.org Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35 Thanks Franco, Your (first) solution is the only one which has worked so far although it utilizes a lambda expression. The problem is indeed tricky. ----------------------------------------------------------------- From: "franco00@gmx.com" To: beginners@haskell.org Sent: Wednesday, March 28, 2012 3:39 PM Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35 gah sorry I obviously meant to reply to the "Unique integers in a list" message ----- Original Message ----- From: franco00@gmx.com Sent: 03/28/12 09:36 AM To: beginners@haskell.org Subject: Re: Beginners Digest, Vol 45, Issue 35 unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = (unique . filter (/= x)) xs | otherwise = x : unique xs -- This is a simpler to read version (albeit inefficient?) unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners