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Date: Sat, 12 Dec 2009 01:33:18 +0000 (UTC)
From: "Bogdan Pistol (JIRA)" <jira@apache.org>
To: dev@mina.apache.org
Subject: [jira] Commented: (DIRMINA-751) IoBuffer.normalizeCapacity
 improvement
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    [ https://issues.apache.org/jira/browse/DIRMINA-751?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=12789643#action_12789643 ] 

Bogdan Pistol commented on DIRMINA-751:
---------------------------------------

using Emmanuel's raw code suggestion passes the tests:

                if (requestedCapacity < 0) {
			return Integer.MAX_VALUE;
		}

		int newCapacity = Integer.highestOneBit(requestedCapacity);
		newCapacity <<= (newCapacity < requestedCapacity ? 1 : 0);
		return (newCapacity < 0 ? Integer.MAX_VALUE : newCapacity);

> 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, 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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