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    <p>If you instrument the code you will see that when you call <code>Derived d2(d1)</code> the Derived::Derived(Base&amp;) constructor is not being called. This is because the d1 argument is a better match for the implicit copy constructor, which just copies the b member from d1 to d2.</p> <p>In order to see the behavior you expect, you can explicitly cast the d1 to <code>(Base&amp;)d1</code>. If you do so you will get code like the following (with the instrumentation):</p> <pre><code>#include &lt;iostream&gt; #include &lt;string&gt; class Base { public: Base() {} virtual ~Base() {} virtual std::string text() const { return "Base"; } }; class Derived: public Base { public: Derived(Base&amp; _b): b(_b) {std::cout &lt;&lt; "init'ed with: " &lt;&lt; _b.text() &lt;&lt; std::endl;} virtual ~Derived() {} virtual std::string text() const { return b.text() + " - Derived"; } private: Base&amp; b; }; int main(int argc, char const *argv[]) { std::cout &lt;&lt; "Creating Base" &lt;&lt; std::endl; Base b; std::cout &lt;&lt; "Creating d1" &lt;&lt; std::endl; Derived d1(b); std::cout &lt;&lt; d1.text() &lt;&lt; std::endl; std::cout &lt;&lt; "Creating d2" &lt;&lt; std::endl; Derived d2(d1); std::cout &lt;&lt; d2.text() &lt;&lt; std::endl; std::cout &lt;&lt; "Creating d3" &lt;&lt; std::endl; Derived d3((Base&amp;)d1); std::cout &lt;&lt; d3.text() &lt;&lt; std::endl; return 0; } </code></pre> <p>And this gives the expected output:</p> <pre><code>Creating Base Creating d1 init'ed with: Base Base - Derived Creating d2 Base - Derived Creating d3 init'ed with: Base - Derived Base - Derived - Derived </code></pre>
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    1. COUgh. Another reason not to have a D(B&) constructor!
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