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Date: Wed, 16 Jan 2013 15:54:15 +0000 (UTC)
From: "Matteo Bertozzi (JIRA)" <jira@apache.org>
To: issues@hbase.apache.org
Message-ID: <JIRA.12627784.1358322827447.145608.1358351655331@arcas>
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Subject: [jira] [Commented] (HBASE-7579) HTableDescriptor equals method
 fails if results are returned in a different order
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    [ https://issues.apache.org/jira/browse/HBASE-7579?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=13555144#comment-13555144 ] 

Matteo Bertozzi commented on HBASE-7579:
----------------------------------------

Add a unit test, to verify equals() and compareTo().

{code}
     result = families.size() - other.families.size();
+      
+      if(result!=0 ) {
+        return result;
+      }
     }
     if (result == 0 && families.size() != other.families.size()) {
{code}
There're lots of redundant check here... 
why we check two times for families.size()?
why we check for == 0 when we return on != 0?


> HTableDescriptor equals method fails if results are returned in a different order
> ---------------------------------------------------------------------------------
>
>                 Key: HBASE-7579
>                 URL: https://issues.apache.org/jira/browse/HBASE-7579
>             Project: HBase
>          Issue Type: Bug
>          Components: Admin
>            Reporter: Aleksandr Shulman
>            Assignee: Aleksandr Shulman
>            Priority: Minor
>             Fix For: 0.96.0, 0.94.5
>
>         Attachments: HBASE-7579-v1.patch
>
>
> HTableDescriptor's compareTo function compares a set of HColumnDescriptors against another set of HColumnDescriptors. It iterates through both, relying on the fact that they will be in the same order.
> In my testing, I may have seen this issue come up, so I decided to fix it.
> It's a straightforward fix. I convert the sets into a hashset for O(1) lookups (at least in theory), then I check that all items in the first set are found in the second.
> Since the sizes are the same, we know that if all elements showed up in the second set, then they must be equal.

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