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POc++ inheritance pointer
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  1. POc++ inheritance pointer
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    copied!<p>it's pretty diffecult for me to describe my problem. I have two classes, I would say <code>Base_A</code> and <code>Derived_A</code>. You can see from the names, the class <code>Derived_A</code> is derived from <code>Base_A</code>. Also in my program I have other two classes <code>Base_B</code> and <code>Derived_B</code> (also with inheritance). The class <code>Base_A</code> contains the object of <code>Base_B</code>, and the class <code>Derived_A</code> contains the object of <code>Derived_B</code>. </p> <pre><code>class Base_A { public: Base_A() {} virtual ~Base_A() {} Base_B b_; Base_B* pointer_; void init() { b_ = Base_B(); pointer_ = &amp;b_; pointer_-&gt;setValue(1); } void print() { pointer_-&gt;getValue(); } }; class Derived_A: public Base_A { public: Derived_A() {} virtual ~Derived_A() {} Derived_B b_; Derived_B* pointer_; void init() { b_ = Derived_B(); pointer_ = &amp;b_; pointer_-&gt;setValue(2); pointer_-&gt;increaseValue(); } }; class Base_B { public: Base_B() {} virtual ~Base_B() {} int value_; void setValue(int value) { value_ = value; } void getValue() { cout &lt;&lt; "Base_B: " &lt;&lt; value_ &lt;&lt; endl; } }; class Derived_B: public Base_B { public: Derived_B() {} virtual ~Derived_B() {} void increaseValue() { value_++; } }; int main() { Derived_A derived_A = Derived_A(); derived_A.init(); derived_A.print(); return 0; } </code></pre> <p>How you can see every class of A has one object of class B and pointer to this object. My problem is, when I call the function <code>print()</code>, it does not take <code>Derived_B* pointer_</code>, but try to access <code>Base_B* pointer_</code>, which is not exist. How I can say in my program, that it should take the pointer according to the class? Or do I need to declarate the <code>Base_B* pointer_</code> inside the <code>Derived_A</code> class like:</p> <pre><code>Base::pointer_ = pointer_; </code></pre> <p>Maybe is there other method or algorithm for my problem?</p> <p>Thank you a lot.</p>
 

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