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    <p>This actually illustrates one of the coolest and most useful features of C++: Temporary variables. Since you specified that the string reference is <code>const</code>, the compiler allows you to pass a reference to a temporary value to that function. So, here's what's happening behind the scenes with <code>Dude d2 = Dude("bye");</code>:</p> <ul> <li>The compiler determines that the best constructor to use is <code>Dude::Dude(const string &amp;)</code>. How this choice is made is a whole different topic.</li> <li>However, in order to use that constructor you need a <code>string</code> value. Now, <code>"bye"</code> is a <code>const char[4]</code>, but the compiler can trivially convert that to a <code>const char *</code>, and <em>that</em> can be turned into a <code>string</code>. So, an anonymous temporary variable (call it <code>temp1</code>) is created.</li> <li><code>string::string(const char *)</code> is invoked with <code>"bye"</code>, and the result is stored in <code>temp1</code></li> <li><code>Dude::Dude(const string&amp;)</code> is invoked with a reference to <code>temp1</code>. The result is assigned to <code>d2</code> (actually, it is assigned to another temporary variable and the copy constructor for <code>Dude</code> is invoked with a const reference to it and <em>that</em> is assigned to d2. But in this case the result is the same.)</li> <li><code>temp1</code> is discarded. This is where the string destructor <code>string::~string()</code> is run on <code>temp1</code></li> <li>Control passes to the next statement</li> </ul>
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    1. USDavid O'Riva
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    1. CO"bye" is not a `const char *`, it is a `const char[4]`. It *decays* to a pointer in the above context.
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      1. USPraetorian
    2. COA subtle but important distinction, especially for someone learning the language who actually wants to know how the type system works. Answer edited to include that.
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