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    <p>That is the way the language used to work. Prior to the <em>using</em> keyword, if you overrode one overloaded function, you had to overload them all:</p> <pre><code>class bar : public foo { public: bar(void); ~bar(void); a(int); a(double d) { foo::a(d); } // add this } </code></pre> <p>This annoyed enough people that the language committee added the <em>using</em> feature, but some old habits die hard; and the habitués&dagger; have a good argument.</p> <p>As James Hopkins points out, by adding <em>using</em>, the programmer is expressing the intention that the derived class will, without warning, add any future overrides of foo::a() to its list of acceptable signatures. </p> <p>Here is an example of what he describes:</p> <pre><code>#include &lt;iostream&gt; class Base { public: virtual void f(double){ std::cout &lt;&lt; "Base::Double!" &lt;&lt; std::endl; } // virtual void f(int) { std::cout &lt;&lt; "Base::Int!" &lt;&lt; std::endl; } // (1) virtual ~Base() {} }; class Derived : public Base { public: // using Base::f; // (2) void f(double) { std::cout &lt;&lt; "Derived::Double!" &lt;&lt; std::endl; } }; int main(int, char **) { Derived d; d.f(21); return 0; } </code></pre> <p>The output will be "Derived::Double!" because the compiler will promote the integer argument to a double. g++ 4.0.1 -Wall will not warn that this promotion occurred.</p> <p>Uncomment (1) to simulate a future change to Base adding the method Base::f(int). The code compiles, again without warning even with -Wall, and "Derived::Double!" remains the output.</p> <p>Now uncomment (2) to simulate a decision by the Derived programmer to include all Base::f signatures. The code compiles (without warnings), but the output is now "Base::Int!".</p> <p>&mdash;</p> <p>&dagger; I cannot think of an English word for "those who have the habit" and "addicted" is much too strong.</p>
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    1. POOverriding a Base's Overloaded Function in C++
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    1. USThomas L Holaday
    1. USThomas L Holaday
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    1. COThis is not true at all, you can very well override just one function.
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      1. USPaolo Tedesco
    2. COMy only defense is that I wrote a lot of C++ before 1996. Sigh.
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    3. COAh, the good old times :)
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