Call derived class method from base class reference

Question

class Material
{
public:
 void foo()
 {
  cout << \"Class Material\";
 }
};

class Unusual_Material : public Material
{
public:
 void foo()
 {
  cout <<          


        

Answer 1

What you want is polymorphism, and to enable it for a function you need to make it virtual:

class Material 
{ 
public: 
    virtual void foo() // Note virtual keyword!
    { 
        cout << "Class Material"; 
    } 
}; 

class Unusual_Material : public Material 
{ 
public: 
    void foo() // Will override foo() in the base class
    { 
        cout << "Class Unusual_Material"; 
    } 
}; 

Also, polymorphism only works for references and pointers:

int main()  
{  
    Unusual_Material unusualMaterial;
    Material& strange = unusualMaterial;
    strange.foo();  
    return 0; 
}

/* OR */

int main()  
{  
    Unusual_Material unusualMaterial;
    Material* strange = &unusualMaterial;
    strange->foo();  
    return 0; 
}

What you have in your code snippet will slice the Unusual_Material object:

int main() 
{ 
    // Unusual_Material object will be sliced!
    Material strange = Unusual_Material(); 
    strange.foo(); 
    return 0; 
} 

Answer 2

Still better explanation would be..

class Base
{
public:
 void foo()     //make virtual void foo(),have derived method invoked
 {
  cout << "Class Base";
 }
};
class Derived: public Base
{
public:
 void foo()
 {
  cout << "Class Derived";
 }
};
int main()
{
 Base base1 = Derived ();
 Base1.foo(); //outputs "Class Base" 
           // Base object, calling base method

 Base *base2 = new Derived ();
 Base2->foo(); //outputs"Class Base",Again Base object calling base method

// But to have base object, calling Derived method, following are the ways
// Add virtual access modifier for base foo() method. Then do as below, to //have derived method being invoked.
//
// Base *base2 = new Derived ();
// Base2->foo();    //outputs "Class Derived" .

return 0;
}