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 )

}

The time complexity of this is O, 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.