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copied!<p>Virtual function calling mechanisms are resolved at run-time. Because, in <strong>C++</strong>, a pointer to a derived class is type-compatible with a pointer to its base class. So, to call the virtual function, the actual type of constructed object( or the actual under lying object ) to which base class pointer is pointing to must be known, which can only be resolved at runtime.</p> <pre><code>struct foo { virtual void virtualMethod() { cout<< " \n virtualMethod of foo \n"; } void normalMethod() { cout<< " \n normalMethod of foo \n"; } virtual ~foo() {} }; struct bar: public foo { void virtualMethod() { cout<< " \n virtualMethod of bar \n"; } void normalMethod() { cout<< " \n normalMethod of bar \n"; } ~bar() {} }; foo* obj = new bar ; obj->virtualMethod() ; </code></pre> <p>Now, since which <code>virtualMethod()</code> needs to be called depends on the run time type( or the actual under lying object ) <code>obj</code> is pointing to because <code>obj</code> can be pointed to an object constructed by either <code>new foo</code> or <code>new bar</code>. At run-time, we know that that <code>obj</code> is constructed from an objected whose type returned is <code>bar*</code>, corresponding virtual function of the derived class is called, if exists. Else base class virtual function is called.</p> <pre><code>obj->normalMethod(); </code></pre> <p>This method can be resolved at compile time itself because it's a normal member function.</p> <p>Results: <a href="http://ideone.com/Uev3P" rel="nofollow">ideOne results link</a></p>

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