C plus plus Interview Questions with Answers Page III

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1.Find the output of the following program
class Sample
{
public:
        int *ptr;
        Sample(int i)
        {
        ptr = new int(i);
        }
        ~Sample()
        {
        delete ptr;
        }
        void PrintVal()
        {
        cout << "The value is " << *ptr;
        }
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}

Answer:
Say i am in someFunc
Null pointer assignment(Run-time error)

Explanation:
As the object is passed by value to SomeFunc  the destructor of the object is called when the control 
returns from the function. So when PrintVal is called it meets up with ptr  that has been freed.The 
solution is to pass the Sample object  by reference to SomeFunc:
 
void SomeFunc(Sample &x)
{
cout << "Say i am in someFunc " << endl;
}
because when we pass objects by refernece that object is not destroyed. while returning from the function.

Which is the parameter that is added to every non-static member function when it is called?
Answer:
            ‘this’ pointer

2.Find the output of the following program
class base
        {
        public:
        int bval;
        base(){ bval=0;}
        };
 
class deri:public base
        {
        public:
        int dval;
        deri(){ dval=1;}
        };
void SomeFunc(base *arr,int size)
{
for(int i=0; i
        cout<bval;
cout<
}
 
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
}
 
Answer:
 00000
 01010

Explanation:  
The function SomeFunc expects two arguments.The first one is a pointer to an array of base class 
objects and the second one is the sizeof the array.The first call of someFunc calls it with an 
array of bae objects, so it works correctly and prints the bval of all the objects. When Somefunc 
is called the second time the argument passed is the pointeer to an array of derived class objects 
and not the array of base class objects. But that is what the function expects to be sent. So the 
derived class pointer is promoted to base class pointer and the address is sent to the function. 
SomeFunc() knows nothing about this and just treats the pointer as an array of base class objects. 
So when arr++ is met, the size of base class object is taken into consideration and is incremented
by sizeof(int) bytes for bval (the deri class objects have bval and dval as members and so is of 
size >= sizeof(int)+sizeof(int) ).

3.Find the output of the following program
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.

4.Find the output of the following program
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.

5.Find the output of the following program
void main()
{
            int a, *pa, &ra;
            pa = &a;
            ra = a;
            cout <<"a="<<<"*pa="<<*pa <<" ra"<}
<<"*pa="<<*pa <<" ra"<
Answer :
            Compiler Error: 'ra',reference must be initialized
<<"*pa="<<*pa <<" ra"<
Explanation :
            Pointers are different from references. One of the main
differences is that the pointers can be both initialized and assigned,
whereas references can only be initialized. So this code issues an error.

6.Find the output of the following program<<"*pa="<<*pa <<" ra"<
const int size = 5;
void print(int *ptr)
{
            cout<
}
 
void print(int ptr[size])
{
            cout<
}
 
void main()
{
            int a[size] = {1,2,3,4,5};
            int *b = new int(size);
            print(a);
            print(b);
}

Answer:
            Compiler Error : function 'void print(int *)' already has a body
 
Explanation:
            Arrays cannot be passed to functions, only pointers (for arrays, base addresses)
can be passed. So the arguments int *ptr and int prt[size] have no difference 
as function arguments. In other words, both the functoins have the same signature and
so cannot be overloaded.

7.Find the output of the following program<<"*pa="<<*pa <<" ra"< 
class some{
public:
            ~some()
            {
                        cout<<"some's destructor"<
            }
};
 
void main()
{
            some s;
            s.~some();
}

Answer:
            some's destructor
            some's destructor
<<"*pa="<<*pa <<" ra"<
Explanation:
            Destructors can be called explicitly. Here 's.~some()' explicitly calls the
destructor of 's'. When main() returns, destructor of s is called again,
hence the result.

8.Find the output of the following program<<"*pa="<<*pa <<" ra"< 
#include
 
class fig2d
{
            int dim1;
            int dim2;
 
public:
            fig2d() { dim1=5; dim2=6;}
 
            virtual void operator<<(ostream & rhs);
};
 
void fig2d::operator<<(ostream &rhs)
{
            rhs <dim1<<" "<dim2<<" ";
}
 
/*class fig3d : public fig2d
{
            int dim3;
public:
            fig3d() { dim3=7;}
            virtual void operator<<(ostream &rhs);
};
void fig3d::operator<<(ostream &rhs)
{
            fig2d::operator <<(rhs);
            rhs<dim3;
}
*/
 
void main()
{
            fig2d obj1;
//          fig3d obj2;
 
            obj1 << cout;
//          obj2 << cout;
}

Answer :
            5 6
<<"*pa="<<*pa <<" ra"<
Explanation:
            In this program, the << operator is overloaded with ostream as argument.
This enables the 'cout' to be present at the right-hand-side. Normally, 'cout'
is implemented as global function, but it doesn't mean that 'cout' is not possible
to be overloaded as member function.
    Overloading << as virtual member function becomes handy when the class in which
it is overloaded is inherited, and this becomes available to be overrided. This is as opposed
to global friend functions, where friend's are not inherited.

9.Find the output of the following program<<"*pa="<<*pa <<" ra"< 
class opOverload{
public:
            bool operator==(opOverload temp);
};
 
bool opOverload::operator==(opOverload temp){
            if(*this  == temp ){
                        cout<<"The both are same objects\n";
                        return true;
            }
            else{
                        cout<<"The both are different\n";
                        return false;
            }
}
 
void main(){
            opOverload a1, a2;
            a1= =a2;
}
 
Answer :
            Runtime Error: Stack Overflow
<<"*pa="<<*pa <<" ra"<
Explanation :
            Just like normal functions, operator functions can be called recursively. This program just 
illustrates that point, by calling the operator == function recursively, leading to an infinite loop.