Question 17: Predict the output.

Question 17: Predict the output. (Considering int for 4 bytes)

class BaseClass
{
public:
    int m_BaseData;
};

class DeriveClass : public BaseClass
{
public:
    int m_DeriveData;
    void printValues()
    {
        std::cout<<"\nBase Data : "<<m_BaseData;

        std::cout<<"\tDerive Data : "<<m_DeriveData;
    }
};

void BaseClassLoop (BaseClass *List, int Count)
{
    int i;

    for (i = 0; i < Count; ++i)
        List[i].m_BaseData = i;

}

int main()
{
    DeriveClass objList[2];
    BaseClassLoop(objList,2);
    for(int i =0;i< 2;i++)
    {
        objList[i].printValues();
    }
}

Options:

(a) Base Data: 0                                Derive Data: Garbage Value

      Base Data: 1                                Derive Data: Garbage Value

(b) Base Data: Garbage Value        Derive Data: Garbage Value

      Base Data: Garbage Value        Derive Data: Garbage Value

(c) Base Data: 0                                 Derive Data: 1

      Base Data: Garbage Value        Derive Data: Garbage Value

(d) None of the above.

Show Solution

Solution: c

Explanation :When object array is passed as an argument, a pointer to array is passed.
On incrementing the pointer it will increment as per the size of Base class that is by one integer instead of two, due to which next time when for loop executed the next memory location value is modified.