Puzzle code of Ponter and virtual function in C++

Q1) Output of the program
#include "iostream.h"
#include "conio.h"
class  abc
{
 private:int a;
 public :void getdata()
                {
                                 a=100;
                };
                void putdata()
                {
                                cout<<a;
                };
};
main()
{
 abc p;
 abc *q=&p;
 q->getdata();
 q->putdata();
 getch();
}
Ans:-100
Discussion:-Here q is a pointer of abc class type .And it takes a address of abc class object.

Q2)Output of the program
#include "iostream.h"
 #include "conio.h"
 class test
 {
                private :int a;
                               int b;

 };

 main()
 {
  test t;
  test *p;
  p=&t;
  p->a=100;
  p->b=200;
  getch();
 }

Ans:-Compile time error.
Discussion:-Here  a  and  b are private data member so, It is not possible to access the data member using pointer directly.


Q3) Output of the program.
#include "iostream.h"
 #include "conio.h"
 class test
 {
                private :int a;
                            int b;
                            test *p;
                public :void getdata()
                            {
                                    test t;
                                     p=&t;
                                     p->a=100;
                                     p->b=200;
                            };
                            void putdata()
                            {
                                      cout<<p->a;
                                       cout<<p->b;
                            }

 };

 main()
 {
  test t;
  t.getdata();
  t.putdata();
  getch();
 }
Ans:-100  200
Discussion:-Here p is a pointer of test class ,and it is declared in private part of the class definition.
                     As p is a pointer it is capable of hold the adress of a class object. And we can access class member using ->(Arrow) operator using pointer.


Q4)Output of the program.
#include "iostream.h"
 #include "conio.h"
 class test
 {
                private :int a;
                            int b;
              public :void getdata()
                            {
                                  a=10;
                                   b=10;
                             };
                           void putdata()
                             {
                                  cout<<a;
                                   cout<<b;
                             }

 };

 main()
 {
  test t;
  test *ptr=new test;                          //Pointer declaration using new operator.
  ptr->getdata();
  ptr->putdata();
  getch();
 }

Ans :- 10  10
Discussion:-Here pointer is declared using new operator.



Q5) Output of the program.
#include "iostream.h"
#include "conio.h"
class base
 {
                public:void display()
                          {
                                cout<<"Base class";
                          };


 };

 class derive: public base
 {
  public:void display()
                {
                         cout<<"Derive class";
                };
 };

 main()
 {
  base b;
  derive d;
  base *p=&b;
  p->display();


  p=&d;
  p->display();    

  getch();
 }
Ans:-Base class.
Discussion:-Here base and derive class containing same display() function signature.  So  when the derive class display() function will call ,the compiler just  neglect the call and compilat just call to the function according to pointer, ie  in which class the pointer is declared.


Q6)Output of the program.
#include "iostream.h"
#include "conio.h"
class base
 {
                public:virtual void display()
                          {
                                 cout<<"Base class";
                          };
 
 };

 class derive:public base
 {
  public:void display()
                {
                                 cout<<"Derive class";
                };

 };

 main()
 {
  base b;
  derive d;
  base *p=&b;
  p->display();


  p=&d;
  p->display();


  getch();
 }
Ans:-Base class  Derived class.
Discussion:-Here in base class the display() function is declared as virtual function. So the display() function will call according to initialization of object into pointer.


Q7) Output of the program
#include "iostream.h"
#include "conio.h"
class base
 {
 public:void print()
           {
                   cout<<"Hello";
           };
  };

 main()
 {
  base b[10];
  base *p=&b[0];
  p->print();
  getch();
 }
Ans:-Hello
Discussion:-Here array of object has created.  And p pointer  holding the address of b[0] object .

Q8)Output of the program
#include "iostream.h"
#include "conio.h"
class test
{
 public:void hello()
           {
                           cout<<"Hello";
           };
};

main()
{
  test t;
  test *ptr;
  (*ptr).hello();              //Hello function call.
  getch();
}
Ans:-Hello.
Discussion:-Hello function is invoking using pointer of the test class.



Q9)Output of the program.

#include "iostream.h"
#include "conio.h"
class abc
{
 public :int a;
           void print()
            {
                     cout<<this->a;
             };
};

main()
{
 abc t;
 t.a=100;
 t.print();
 getch();
}

output:-100
discussion:-This pointer always point current object member.

Q10)Output of the program.

#include "iostream.h"
#include "conio.h"
class abc
{
 public :void print()
            {
                    cout<<"I am base function";
            };
};
class xyz:public abc
{
                public :void print()
                           {
                                            cout<<"I am base function";
                           };
};

main()
{
 xyz *p;
 abc a;
 p=&a;
 getch();
}
ans:-Error
Discussion:-Here the derive class pointer is trying to assign with the base class object.

Q11)Output of the program
#include "iostream.h"
#include "conio.h"
class abc
{
 public :virtual void print()=0;

};
class xyz:public abc
{
                public:void print()
                                                {
                                                 cout<<"I am derive function";
                                                };
};

main()
{
 xyz t,*p;
 p=&t;
 p->print();
 getch();
}
Output:-I am derive function
Discussion:-Here in base class the print function is declared as pure virtual function.

Q12)Output of the program.
#include "iostream.h"
#include "conio.h"
class abc
{
  private:int a,b;
                 void hello()
                 {
                  cout<<"I am hello function";
                 };

};

main()
{
 abc *t=new abc();
 cout<<sizeof(*t);
 getch();
}
Output:-4 (in 32 bit turbo c++ compilar)
Discussion:-Output is compilar dependent.Here two int data are declared in abc class.So result is 4.

Q13)Output of the program
#include "iostream.h"
#include "conio.h"

main()
{
 int *p=new int[10];

 *p=10;
 cout<<*p;
 getch();
}
Ans:-10
Discussion:-Here new operator creates 10*2=10 byte memory and initialize the base adress into p.

Q14)Output of the program
#include "iostream.h"
#include "conio.h"
class string
{
 private:char *p;
 public :char * hello(char *t)
                {
                                p=t;
                                return p;
                };
};
main()
{
 string s;
 char *c=s.hello("Hello world");
 cout<<c;
 getch();
}
Ans:-Hello world
Discussion:-Here string pointer is passing as a hello function argument and the hello function is returning the string pointer again in main function.

Complex code of operator overloading in C++

#include "iostream.h"
#include "conio.h"
class test
{
  private:int a;
                 int b;
  public:void getdata()
                {
                                 a=10;
                                 b=10;
                };
                void putdata()
                {
                                 cout<<a;
                                 cout<<b;
                };
                void operator-()
                {
                                 a=-a;
                                 b=-b;
                };


};
main()
{
test t;
t.getdata();
-t;
t.putdata();
 getch();
}
Ans:-    -10    -10
Discussion:-Here –(Ve) operator is overloading. And it changes the sign of a and b.


Q2)Output of the program
#include "iostream.h"
#include "conio.h"
class test
{
  private:int a;
                 int b;
  public:void getdata()
                {
                                 a=10;
                                 b=10;
                };
                void putdata()
                {
                               cout<<a;
                                 cout<<b;
                };
                void operator::()
                {
                                 a=-a;
                                 b=-b;
                };


};

main()
{
test t;
t.getdata();
::t;
t.putdata();
 getch();
}
Ans:-Compile time error.
Discussion:-Here ::  (Scope resolution) operator is trying to overload. But it not possible to overload scope resolution operator.


Q3)Output of the program.
#include "iostream.h"
#include "conio.h"
class test
{
  private:int a;
                 int b;
  public:void getdata()
                {
                                 a=10;
                                 b=10;
                };
                void putdata()
                {
                                 cout<<a;
                               cout<<b;
                };
                test operator+(test t)
                {
                                 test temp;
                                 temp.a=a+t.a;
                                 temp.b=b+t.b;
                                 return temp;
                };


};

main()
{
test t1,t2,t3;
t1.getdata();
t2.getdata();
t3=t1+t2;
t3.putdata();
 getch();
}
Ans:-20 20
Discussion:-Here binary (+) operator is overloaded. And It adding two object  t 1 and t2.

Puzzle code of inheritence in C++

Q1)Output of the program.
#include "iostream.h"
#include "conio.h"
class base
{
 private:void print()
                {
                                  cout<<"Base class";
                }

};
class derive:public base
{
 public:void call()
                  {
                                 cout<<"Derive class";
                  }
}

main()
{
derive d;
d.print();
 getch();
}
Ans:-Compile time error.
Discussion:-Here derive class is publicly inherited from base class. So the print() function of base class remain private member in derive class. So  the print() function call is not possible  using direct object.

Q2)Output of the program.
#include "iostream.h"
#include "conio.h"
class base
{
 private:int a;
 public: void print()
                {
                                  a=100;
                                  cout<<a;
                }


};
class derive:public base
{
 public:void call()
                  {
                                 print();
                  }
};

main()
{
derive d;
d.call();
 getch();
}
Ans-100
Discussion:-Here derive class is publicly inherited from base class. So derive  class function  call() is calling the base class function print().

Q3) Output of the program.
#include "iostream.h"
#include "conio.h"
class base
{
  public: base()
                  {
                                 cout<<"Base class constructor"<<endl;
                  }


};
class derive:public base
{
 public:derive()
                 {
                                 cout<<"Derive class constructor"<<endl;
                                 base();
                 }

};

main()
{
derive d;
getch();
}
Ans:-Base class constructor.
         Derive  class constructor.
         Base  class constructor.

Discussion:-When the constructor of derive class is call, then at first the base class constructor is called then sub class constructor is call.

Q4)Output of the program
#include "iostream.h"
#include "conio.h"
class base
{
  public:void abc()
                                  {
                                                 cout<<"Base class"<<endl;
                                  };


};
class derive:private base
{
 public:void xyz()
                                 {
                                                abc();
                                 };

};

main()
{
derive d;
d.abc();
getch();
}
Ans:-Compile time error.
Discussion:-Here derive class is privately inherited from base class.  So  public member of base class will be the private member of sub class.

Q5)Output of the program
#include "iostream.h"
#include "conio.h"
class base
{
  public:void abc()
              {
                                 cout<<"Base class"<<endl;
                 };


};
class derive:private base
{
 public:void xyz()
                 {
                                abc();
                 };
};
class low:public derive
{
                  public:void hello()
                  {
                                  xyz();
                 }

};
main()
{
low l;
l.hello();
getch();
}
Ans:-Base class.
Discussion:-It is the example of multi level inheritance. Her e hello() function is calling it’s superclass xyz() function and xyz() function is calling it’s super class  function abc().And within abc() function ,the message is printing.

Q6)Output of the program.
#include "iostream.h"
#include "conio.h"
class base
{
  protected:void abc()
                      {
                                 cout<<"Base class"<<endl;
                      };


};
class derive:private base
{
 public:void xyz()
                 {
                                abc();
                 };
};

main()
{
derive d;
d.xyz();
getch();
}
Ans:-Base class.
Discussion:-Here derive class is privately inherited from base class. So protected member of base class will be the private member of derive class.


Q7)Output of the program.
#include "iostream.h"
#include "conio.h"
class base
{
  protected:void abc()
                      {
                                cout<<"Base class"<<endl;
                      };


};
class derive:public base
{
 public:void xyz()
                 {
                                abc();
                 };
};

main()
{
derive d;
d.abc();
getch();
}
Ans:-Compile time error.
Discussion:-Here derive class is inherited publicly. So the protected function member abc() remain protected in derive class. So it is not possible to access the protect  function member using  . operator.


Q8) Output of the program.
#include "iostream.h"
#include "conio.h"
class abc
{
 private  :void abcfun()
                  {
                                cout<<"Super class1"<<endl;
                  };
 public:  void callabc()
                 {
                              abcfun();
                 }
};
class xyz:public abc
{
 public:void xyzfun()
                 {
                                cout<<"Super class2";
                              callabc();

                };
};
class pqr:private abc,public xyz
{
 public :void pqrfun()
                {
                                 cout<<"Derive class";
                }
};

main()
{
 pqr p;
 p.pqrfun();
 p.xyzfun();
 getch();
}

Ans:-Derive class Super class2 Super class1
Discussion:- It is a example of hierarchical inheritance. Here xyz class is publicly inherited from  base class abc and class pqr is inherited both from abc as privately and xyz  as publicly.


Q9)Output of the program.
#include "iostream.h"
#include "conio.h"
class abc
{
 private:int a;
 public:abc()
            {
                                cout<<"Hello";
             };

};

class xyz:public abc
{
 private:int x;
 public:xyz()
             {
                                abc();
             };

};

class pqr:public xyz,public abc
{
 public:pqr()
                 {
                                xyz();
                  }
};

main()
{
 pqr p;

 getch();
}
Ans:-HelloHelloHelloHello


Q10)Output of the program
#include "iostream.h"
#include "conio.h"
class abc
{
                public:void print()
                            {
                                  cout<<"abc";

                             };
};

class xyz
{
 public:void print()
                 {
                                  cout<<"xyz";

                 };


};

class pqr: public xyz,public abc
{


};
main()
{
 pqr p;
 p.abc::print();                   //Call to print function of abc class
 p.xyz::print();                   // Call to print function of xyz class

 getch();
}
Ans:-abc  xyz
Discussion:-Here pqr  class is publicly inherited from both the class abc and xyz.
                      And is print() function is common both two class ,so we can invoke individual print() function using scope resolution operator.