C++

1) class base

        {

        public:

            void baseFun(){ cout<<"from base"<

        };

 class deri:public base

        {

        public:

            void baseFun(){ cout<< "from derived"<

        };

void SomeFunc(base *baseObj)

{

        baseObj->baseFun();

}

int main()

{

base baseObject;

SomeFunc(&baseObject);

deri deriObject;

SomeFunc(&deriObject);

}

Answer:

            from base

            from base

Explanation:

            As we have seen in the previous case, SomeFunc expects a pointer to a base class. Since a pointer to a derived class object is passed, it treats the argument only as a base class pointer and the corresponding base function is called.

2) class base

        {

        public:

            virtual void baseFun(){ cout<<"from base"<

        };

 class deri:public base

        {

        public:

            void baseFun(){ cout<< "from derived"<

        };

void SomeFunc(base *baseObj)

{

        baseObj->baseFun();

}

int main()

{

base baseObject;

SomeFunc(&baseObject);

deri deriObject;

SomeFunc(&deriObject);

}

Answer:

            from base

            from derived

Explanation:

            Remember that baseFunc is a virtual function. That means that it supports run-time polymorphism. So the function corresponding to the derived class object is called.