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POHow to replace pointers with references in C++?
 

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  1. POHow to replace pointers with references in C++?
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    <p>"I am sure there are tens of questions with the same title. Many of them are duplicate. Mine might be duplicate too, but I couldn't find any. So I try to make it very neat, short and simple."</p> <p>I have an hierarchy like this:</p> <pre><code>class Shape { public: virtual void virtualfunc() { std::cout &lt;&lt; "In shape\n"; } }; class Circle: public Shape { public: void virtualfunc() { std::cout &lt;&lt; "In Circle\n"; }; }; </code></pre> <p>and when I use the classes with the help of pointer, the functions are called as I expected:</p> <pre><code>int main() { Shape shape_instance; Shape* ref_shape = &amp;shape_instance ; Circle circle_instance; Circle* ref_circle = &amp;circle_instance; ref_shape = dynamic_cast&lt;Shape*&gt; (ref_circle); ref_shape-&gt;virtualfunc(); } </code></pre> <p>Here the program calls the <code>virtualfunc()</code> of the derived class and the result is naturally : <code>In Circle</code> </p> <p>Now, I want to get rid of the pointers, use references instead, and get the same result. So I make trivial modifications to <code>main()</code> to look like this: </p> <pre><code>int main() { Shape shape_instance; Shape&amp; ref_shape = shape_instance; Circle circle_instance; Circle&amp; ref_circle = circle_instance; ref_shape = dynamic_cast&lt;Shape&amp;&gt;(ref_circle); ref_shape.virtualfunc(); } </code></pre> <p>But this time, the program calls the <code>virtualfunc()</code> of the base class and the result is : <code>In Shape</code></p> <p>I appreciate if you let me know which concept of the references I am missing and how to change the references in the main() to get the result I got in the pointer version.</p> <p>thank you</p>
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    1. USreinierpost
    1. USrahman
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    1. COWhy did you declare ref_shape again at this line Shape& ref_shape = dynamic_cast<Shape&>(ref_circle);?
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      1. UStaocp
    2. COsorry, a typo. I will just edit that
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    3. COThere is no need to `dynamic_cast` to the base, that is an implicit conversion.
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      1. USDavid Rodríguez - dribeas
 

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