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2011-01-24T21:25:19.517
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thanks for the answers.. helped me a lot to understand the problem. In effect I posted some wrong code, because i was misunderstanding the real problem. Anyway, i think i partially solved my problem. Here's the code: <pre><code> #include <iostream> `` using namespace std; class Other{ public: void foo(){ cout << "Other\n"; } void foo(int a){} }; class Abstract : public Other{ public: virtual void foo() {}; virtual void foo(int c){ Other::foo(c); } }; class A : public Abstract{ public: void foo(){ cout << "A\n"; } }; class B : public Abstract{ public: void foo(){ cout << "B\n"; } }; int main(){ cout << "main\n"; Abstract * ptrA = new A(); Abstract * ptrB = new B(); Other *o = new Other(); o->foo(); ptrA->foo(); ptrB->foo(); ptrB->foo(3); //can't no more use the method foo with different signatures implemented in the base class Other, unless I explicitly redefined in the class Abstract dynamic_cast<Other*>(ptrB)->foo(3);//can't dynamic_cast from derived to base </code></pre> I was making two errors: <ol> <li>In my real code (not the simplified version posted before) i forgot to declare virtual the function foo()</li> <li>Even declaring virtual wasn't enough. In fact all the implementations of that function must be wrapped inside the class Abstract to become visible to the subclasses A and b. Otherwise wont't compile.</li> </ol> I don't know if it could be a clean solution..in fact that way I need to wrap all foo method signatures. 
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