Hi L.Guo Brent has replied with the right answer. The definition of groupBy is below - the span in the where clause only compares with the first element of the current sub-list: -- | The 'groupBy' function is the non-overloaded version of 'group'. groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs -- list1 = [7,3,5,9,6,8,3,5,4::Int] Here are two more runs on your input with a wee bit less clutter. In both cases there are 'spans' where the numbers increase and decrease - this is because the operation is comparing the rest of the input against the first element in the 'span' not consecutive elements. *GroupBy Data.List> groupBy (<) list1 [[7],[3,5,9,6,8],[3,5,4]] *GroupBy Data.List> groupBy (<=) list1 [[7],[3,5,9,6,8,3,5,4]] Best wishes Stephen

On Mon, Dec 07, 2009 at 12:15:45PM -0500, Brent Yorgey wrote:

On Tue, Dec 08, 2009 at 12:45:59AM +0800, L.Guo wrote:

...

Hi there:

My friend asked me a question, and i suppose he has found a bug of `groupBy'.

Here is the code piece:

...

List.groupBy (\a b -> Foreign.unsafePerformIO (Text.Printf.printf "\t%d <= %d ?: %s\n" a b (show (a<=b)) >> return (a<=b))) [7,3,5,9,6,8,3,5,4]

I have tested it in GHC 6.10.4 (Win XP) and GHC 6.8.3 (Linux), both give the wrong result (categaried):

7 <= 3 ?: False 3 <= 5 ?: True 3 <= 9 ?: True 3 <= 6 ?: True 3 <= 8 ?: True 3 <= 3 ?: True 3 <= 5 ?: True 3 <= 4 ?: True [[7],[3,5,9,6,8,3,5,4]]

This looks like the right result to me. What makes you think it is wrong? What result do you expect?

I just realized I think I understand what the confusion is. You think it should return [[7],[3,5,9],[6,8],[3,5],[4]]? But in fact, groupBy only works that way for equivalence relations, where it makes perfect sense to just compare the first thing in each sublist against all the rest, instead of comparing each pair of adjacent elements. The behavior you want is something I have been planning to add to the Data.List.Split module for a while. -Brent