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    <p>Sounds like this might be useful:</p> <p><a href="http://www.aggregate.org/MAGIC/#Population%20Count%20%28Ones%20Count%29" rel="nofollow">http://www.aggregate.org/MAGIC/#Population%20Count%20%28Ones%20Count%29</a> and <a href="http://www.aggregate.org/MAGIC/#Leading%20Zero%20Count" rel="nofollow">http://www.aggregate.org/MAGIC/#Leading%20Zero%20Count</a></p> <p>You don't say if you wanted to do some sort of RLE or to simply count in-bytes zeros and one bits (like 0b1001 should return 1x1 2x0 1x1).</p> <p>A look up table plus SWAR algorithm for fast check might gives you that information easily. A bit like this:</p> <pre><code>byte lut[0x10000] = { /* see below */ }; for (uint * word = words; word &lt; words + bitmapSize; word++) { if (word == 0 || word == (uint)-1) // Fast bailout { // Do what you want if all 0 or all 1 } byte hiVal = lut[*word &gt;&gt; 16], loVal = lut[*word &amp; 0xFFFF]; // Do what you want with hiVal and loVal </code></pre> <p>The LUT will have to be constructed depending on your intended algorithm. If you want to count the number of contiguous 0 and 1 in the word, you'll built it like this:</p> <pre><code> for (int i = 0; i &lt; sizeof(lut); i++) lut[i] = countContiguousZero(i); // Or countContiguousOne(i) // The implementation of countContiguousZero can be slow, you don't care // The result of the function should return the largest number of contiguous zero (0 to 15, using the 4 low bits of the byte, and might return the position of the run in the 4 high bits of the byte // Since you've already dismissed word = 0, you don't need the 16 contiguous zero case. </code></pre>
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    1. POFast code for searching bit-array for contiguous set/clear bits?
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