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From "Bogdan Pistol (JIRA)" <j...@apache.org>
Subject [jira] Updated: (DIRMINA-751) IoBuffer.normalizeCapacity improvement
Date Sun, 13 Dec 2009 00:43:18 GMT

     [ https://issues.apache.org/jira/browse/DIRMINA-751?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel
]

Bogdan Pistol updated DIRMINA-751:
----------------------------------

    Attachment: patch-lookup-tables.txt
                IoBufferTest.java

Ok, I've written the JUnit test (it's attached to this issue), but I also had another idea
to make this even faster
the new implementation is 60% faster.

I'll stop here with the improvements for normalizeCapacity()   :), I think it's enough.

> IoBuffer.normalizeCapacity improvement
> --------------------------------------
>
>                 Key: DIRMINA-751
>                 URL: https://issues.apache.org/jira/browse/DIRMINA-751
>             Project: MINA
>          Issue Type: Improvement
>          Components: Core
>    Affects Versions: 2.0.0-RC1
>         Environment: N/A
>            Reporter: Bogdan Pistol
>            Priority: Minor
>             Fix For: 2.0.0-RC2
>
>         Attachments: IoBufferTest.java, IoBufferTest.java, IoBufferTest.java, patch-lookup-tables.txt,
patch.txt
>
>
> The technique of computing the minimum power of 2 that is bigger than the requestedCapacity
in the org.apache.mina.core.buffer.IoBuffer.normalizeCapacity() is not optimal.
> The current computation is as follows:
>  int newCapacity = 1;
>         while ( newCapacity < requestedCapacity )  {
>             newCapacity <<= 1;
>             if ( newCapacity < 0 )   {
>                 return Integer.MAX_VALUE;
>             }
>         }
> The time complexity of this is O(n), where n is the number of bits of the requestedCapacity
integer, that is log2(requestedCapacity) - maximum 31.
> This creates an unnecessary overhead in some high IoBuffer allocations scenarios that
are calling IoBuffer.normalizeCapacity() a lot when creating IoBuffers. I observed this when
benchmarking a MINA server with hprof.
> There is a better solution to this problem than to iterate the bits from 0 to log2(requestedCapacity).
> The alternative is to use a binary search technique that has optimal time complexity
of O(5). Because requestedCapacity is an integer and has a maximum of 2^5 (32) bits we can
binary search in the set of bits and determine in O(5) comparisons the minimum power of 2
that is bigger than the requestedCapacity.

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