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    <p>There's no value type in C++ to represent a single bit, so when you use the <code>[]</code> operator to access an element of a bitset, what you get is a <em>proxy object</em> that serves as an alias for the bit you requested. Assigning to that proxy object changes the corresponding bit value in the original bitset object.</p> <p>As <a href="https://stackoverflow.com/questions/2003255/why-the-stdswap-of-bits-in-a-stdbitset-instance-doesnt-work/2003286#2003286">Victor's answer</a> shows, your code doesn't compile with GCC. But let's suppose the call to <code>swap</code> would compile. You'd get code that amounts to something like this:</p> <pre><code>void swap(std::bitset&lt;2&gt;::reference&amp; a, std::bitset&lt;2&gt;::reference&amp; b) { std::bitset&lt;2&gt;::reference tmp = a; a = b; b = tmp; } </code></pre> <p>The <code>tmp</code> declaration initializes the variable with <code>a</code>, but that doesn't make a copy of the bit. Instead, it makes a copy of the proxy object, so <code>tmp</code> <em>refers</em> to the same bit in the same bitset that <code>a</code> refers to. The next line assigns <code>b</code> to <code>a</code>, which copies the bit value from the location <code>a</code> refers to and stores it in the bit location <code>b</code> refers to. Finally, there's the assignment of <code>tmp</code> into <code>b</code>. But remember that <code>tmp</code> still refers to the bit that <code>a</code> refers to. Reading <code>tmp</code> is the same as reading <code>a</code>, so you ultimately get the same effect you'd get if <code>swap</code> were just these two lines:</p> <pre><code>a = b; b = a; </code></pre> <p>In your code, <code>a</code> is 0 and <code>b</code> is 1, so with those two assignment statements, 11 is exactly what we'd expect to see.</p>
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