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    <p>There's a simple trick for this problem:</p> <pre><code>bool IsPowerOfTwo(ulong x) { return (x &amp; (x - 1)) == 0; } </code></pre> <p>Note, this function will report <code>true</code> for <code>0</code>, which is not a power of <code>2</code>. If you want to exclude that, here's how:</p> <pre><code>bool IsPowerOfTwo(ulong x) { return (x != 0) &amp;&amp; ((x &amp; (x - 1)) == 0); } </code></pre> <h3>Explanation</h3> <p>First and foremost the bitwise binary &amp; operator from MSDN definition:</p> <blockquote> <p>Binary &amp; operators are predefined for the integral types and bool. For integral types, &amp; computes the logical bitwise AND of its operands. For bool operands, &amp; computes the logical AND of its operands; that is, the result is true if and only if both its operands are true.</p> </blockquote> <p>Now let's take a look at how this all plays out:</p> <p>The function returns boolean (true / false) and accepts one incoming parameter of type unsigned long (x, in this case). Let us for the sake of simplicity assume that someone has passed the value 4 and called the function like so:</p> <pre><code>bool b = IsPowerOfTwo(4) </code></pre> <p>Now we replace each occurrence of x with 4:</p> <pre><code>return (4 != 0) &amp;&amp; ((4 &amp; (4-1)) == 0); </code></pre> <p>Well we already know that 4 != 0 evals to true, so far so good. But what about:</p> <pre><code>((4 &amp; (4-1)) == 0) </code></pre> <p>This translates to this of course:</p> <pre><code>((4 &amp; 3) == 0) </code></pre> <p>But what exactly is <code>4&amp;3</code>?</p> <p>The binary representation of 4 is 100 and the binary representation of 3 is 011 (remember the &amp; takes the binary representation of these numbers). So we have:</p> <pre><code>100 = 4 011 = 3 </code></pre> <p>Imagine these values being stacked up much like elementary addition. The <code>&amp;</code> operator says that if both values are equal to 1 then the result is 1, otherwise it is 0. So <code>1 &amp; 1 = 1</code>, <code>1 &amp; 0 = 0</code>, <code>0 &amp; 0 = 0</code>, and <code>0 &amp; 1 = 0</code>. So we do the math:</p> <pre><code>100 011 ---- 000 </code></pre> <p>The result is simply 0. So we go back and look at what our return statement now translates to:</p> <pre><code>return (4 != 0) &amp;&amp; ((4 &amp; 3) == 0); </code></pre> <p>Which translates now to:</p> <pre><code>return true &amp;&amp; (0 == 0); </code></pre> <pre><code>return true &amp;&amp; true; </code></pre> <p>We all know that <code>true &amp;&amp; true</code> is simply <code>true</code>, and this shows that for our example, 4 is a power of 2.</p>
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    1. POHow to check if a number is a power of 2
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    1. USGreg Hewgill
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    1. POHow to check if a number is a power of 2
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    1. CONice! So simple and elegant. Why didn't I think of that?
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    2. COlooks awesome. didn't understand it though :)
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    3. CO@Kripp: The number will be of the binary form 1000...000. When you -1 it, it will be of the form 0111...111. Thus, the two number's binary and would result is 000000. This wouldn't happen for non-power-of-twos, since 1010100 for example would become 1010011, resulting in an (continued...)
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