How can a derived class pointer to a base class object call methods of the derived class?

Question

class BaseClass{

public:
    int a;
    char buf[250];

    void abcd(){
        cout<<"hello base from abcd";
    }

    virtual void defg(){
        cout<<"hellow base from defg";
    }
};

class DerivedClass: public BaseClass{

public :
    int b;
    char bufb[255];

    void bcde(){
        cout<<"hello base from bcde";
    }

    virtual void defg(){
        cout<<"hellow base from defg";
    }
};

int main(){

    BaseClass* bas=new BaseClass();
    DerivedClass* der;

    der=static_cast<DerivedClass*>(bas);
    cout<<"address of der ="<<der <<"   base= "<<bas<<endl;
    der->bcde();
    cout<<endl<<"Base size:"<< sizeof(*bas)<<"  Derived size:"<<sizeof(*der)<<endl;

    getch();
    return 0;
}

How can derived class pointer to base class object can call methods of derived class. WHY and HOW ? As base class object created in memory is of only size of base class, how could it contains derived class methods. and it successfully got called.

OutPut:==>

address of der =002585A8 base= 002585A8

hello base from bcde

Base size:260 Derived size:520

Answer 1

By casting the pointer to DerivedClass* , when the most derived object is of type BaseClass , and by then using that pointer to access DerivedClass stuff, you have Undefined Behavior , UB. One possible effect of UB is that what one mistakenly believed would happen, happens. Other possibilities include crashes, hangs and weird results.

In practice, since DerivedClass::bcde() doesn't access any data in the instance, and doesn't call any virtual function, the only possible crash cause would be explicit checking inserted by the compiler.

I do not know of any compiler that adds such checking, but the compiler is free to do so, and generally it's free to just assume that UB will not happen.

Answer 2

The compiler bound the non-virtual method bcde based purely on the type information. As the method doesn't refer to the object contents, memory layout did not come into play.

Answer 3

You have a mistakes in the output messages

class DerivedClass: public BaseClass{

public :
    int b;
    char bufb[255];

    void bcde(){
        cout<<"hello base from bcde";
    }

    virtual void defg(){
        cout<<"hellow base from defg";
    }
};

DerivedClass should print "hello der from bcde", not "hello base from bdce". And actually your code prints message from DerivedClass.