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https://issues.apache.org/jira/browse/DIRMINA-751?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel
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Bogdan Pistol updated DIRMINA-751:
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Attachment: IoBufferTest.java
This is a test that proves the the patch is faster than the current
implementation.
It also includes sanity tests that prove the correctness of the implementation.
I tested it with mina-core-2.0.0-RC2 and I got the following results:
test1 - 6032 milliseconds
test2 - 7014 milliseconds
With mina patched with the binary search the results are:
test1 - 1739 ms
test2 - 1899 ms
You can see that this is faster 3.5 times.
The tests were done on a Laptop with 2GHz Dual-Core
> 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-RC1
>
> 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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