haskell - Recursion error while sorting a list of numbers without quicksort - 


order :: [int] -> [int] order [] = [] order [x] = [x] order (x:y:xs) = if x>y [y] ++ (order (x:xs)) else [x] ++ (order (y:xs))

i tried use haskell code sort list of numbers, when input list more 4 elements doesn't sort them correctly. want keep code compiles don't know how make recursion work correctly.

the code fine , implemented, implemented not in fact sort list. let's try [4,3,2]:

order [4,3,2]  order [] = []               -- nope not empty list order [x] = [x]             -- nope not singleton order (x:y:xs) = if x>y     -- x=4, y=3, xs =[2]; x >y  true [y] ++ (order (x:xs))  -- branch [3] ++ (order [4,2]) else [x] ++ (order (y:xs))  -- not branch

so [3] ++ (order [4,2]). question you: how 2 move other side of 3?


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