Don't shout at me for top posting
In this instance it's justified
Thanks for your continued work on this. I have to get some lines of code
down
as release date is fast approaching but I will try your code as soon as I
have time
Thanks for you continued work on this
Lyallex
On 9 November 2012 05:08, Christopher Schultz
<ch...@christopherschultz.net>wrote:
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> Russ,
>
> On 11/8/12 6:05 PM, Russ Kepler wrote:
> > On Thursday, November 08, 2012 07:36:20 PM Lyallex wrote:
> >
> >> The only difference between the two executions is the fact that
> >> the test code executes in it's own instance of the JVM whereas
> >> the other execution runs in an instance shared with the
> >> container.
> >>
> >> I accept that the behaviour may be undefined even though it is
> >> consistently repeatable in both environments but surely given
> >> everything else being equal the results should be the same ... or
> >> maybe I'm just losing the plot.
> >
> > No, you're right but just missing some small difference in the
> > environments.
> >
> > I'd verify that you get the same input data in the same order in
> > both cases, and that you're starting with the same size container
> > [...]
>
> After writing a bench test that I couldn't get to fail, your comment
> here tripped a thought in my brain: the "container" size. So, I added
> an element to my list of Strings and boom: failure. It turns out that
> the collection size doesn't matter: I just hadn't been iterating
> enough, so I added a loop that will run until the initial sorted order
> doesn't match the re-sorted order (with shuffles in between).
>
> Lyallex, see the code below: it will fail after a few iterations to
> produce the same element ordering. Switch from BrokenSorter to
> WorkingSorter and you'll find that it runs forever.
>
> Are you *sure* that your database always returns the items in the same
> order? If you plan on sorting alphabetically later, why bother sorting
> by id when fetching? Unless you are really sorting by id when
> fetching, the data can come back in any order. It may *often* be in
> entry-sequenced order, but it is certainly not guaranteed to be.
>
> The code below shows that, without any funny business, the sort can
> work sometimes and not in others.
>
> Enjoy,
> - -chris
>
> import java.util.ArrayList;
> import java.util.Arrays;
> import java.util.Collections;
> import java.util.Comparator;
> import java.util.List;
>
> public class SortTest
> {
> public static void main(String[] args)
> {
> String[] fruits = new String[] {
> "Apples",
> "Bananas",
> "Coconuts",
> "Dates",
> "Eggplants",
> "Figs",
> "Grapefruits",
> "Honeydews",
> "Ilamas",
> "Jambolans",
> "Kepels",
> "Lemons",
> "Miscellaneous",
> "Nectarines"
> };
>
> List<String> fruitList = Arrays.asList(fruits);
>
> Comparator<String> sorter = new BrokenSorter();
>
> System.out.println("Initial order: " + fruitList);
>
> Collections.sort(fruitList, sorter);
> System.out.println("Sort 1: " + fruitList);
>
> List<String> saved = new ArrayList<String>(fruitList);
>
> int i = 1;
> do
> {
> Collections.shuffle(fruitList);
> Collections.sort(fruitList, sorter);
> System.out.println("Sort " + (++i) + ": " + fruitList);
> }
> while(fruitList.equals(saved));
> System.out.println("Stopped after " + i + " iterations because
> the list did not sort the same way.");
> }
>
> static class BrokenSorter
> implements Comparator<String>
> {
> @Override
> public int compare(String a, String b)
> {
> if(a.equals("Miscellaneous"))
> return 1;
> return a.compareTo(b);
> }
> }
> static class WorkingSorter
> implements Comparator<String>
> {
> @Override
> public int compare(String a, String b)
> {
> if(a.equals("Miscellaneous"))
> return 1;
>
> if(b.equals("Miscellaneous"))
> return -1;
>
> return a.compareTo(b);
> }
> }
> }
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