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

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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  

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