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From "Hadoop QA (JIRA)" <j...@apache.org>
Subject [jira] [Commented] (HBASE-7579) HTableDescriptor equals method fails if results are returned in a different order
Date Thu, 17 Jan 2013 02:18:22 GMT

    [ https://issues.apache.org/jira/browse/HBASE-7579?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=13555758#comment-13555758
] 

Hadoop QA commented on HBASE-7579:
----------------------------------

{color:red}-1 overall{color}.  Here are the results of testing the latest attachment 
  http://issues.apache.org/jira/secure/attachment/12565236/HBASE-7579-v3.patch
  against trunk revision .

    {color:green}+1 @author{color}.  The patch does not contain any @author tags.

    {color:green}+1 tests included{color}.  The patch appears to include 13 new or modified
tests.

    {color:red}-1 patch{color}.  The patch command could not apply the patch.

Console output: https://builds.apache.org/job/PreCommit-HBASE-Build/4064//console

This message is automatically generated.
                
> 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, HBASE-7579-v2.patch, HBASE-7579-v3.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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