250+ TOP MCQs on Pointers to Members and Answers

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This section on C++ language interview questions and answers on “Pointers to Members”. One shall practice these interview questions to improve their C++ programming skills needed for various interviews (campus interviews, walkin interviews, company interviews), placements, entrance exams and other competitive exams. These questions can be attempted by anyone focusing on learning C++ programming language. They can be a beginner, fresher, engineering graduate or an experienced IT professional. Our C++ language interview questions come with detailed explanation of the answers which helps in better understanding of C++ concepts.

Here is a listing of C++ language interview questions on “Pointers to Members” along with answers, explanations and/or solutions:

1. Which is referred by pointers to member?
a) Static members of class objects
b) Non-static members of class objects
c) Referring to whole class
d) Dynamic members of class objects
Answer: b
Clarification: We cannot use a pointer to member to point to a static class member because the address of a static member is not associated with any particular object.

2. What should be used to point to a static class member?
a) Smart pointer
b) Dynamic pointer
c) Normal pointer
d) Static pointer
Answer: c
Clarification: Normal pointer is sed to point to a static class member.

3. Which operator is used in pointer to member function?
a) .*
b) ->*
c) Both .* & ->*
d) $*
Answer: c
Clarification: The pointer to member operators .* and ->* are used to bind a pointer to a member of a specific class object.

4. What will be the output of the following C++ code?

  1.     #include 
  2.     using namespace std;
  3.     class X 
  4.     {
  5.         public:
  6.         int a;
  7. 		void f(int b) 
  8. 		{
  9. 			cout<< b << endl;
  10. 		}
  11.     };
  12.     int main() 
  13.     {
  14.         int X :: *ptiptr = &X :: a;
  15.         void (X :: * ptfptr) (int) = &X :: f;
  16.         X xobject;
  17.         xobject.*ptiptr = 10;
  18.         cout << xobject.*ptiptr << endl;
  19.         (xobject.*ptfptr) (20);
  20.     }

a) 

   10
   20

b) 

   20
   10

c) 20
d) 10
Answer: a
Clarification: In this program, We are assigning 10 and printing it in the
main function and then for value 20, We are passing the value to class and
printing it.
Output:

$ g++ ptm.cpp
$ a.out
10
20

5. What will be the output of the following C++ code?

  1.     #include 
  2.     using namespace std;
  3.     class Testpm 
  4.     {
  5.         public:
  6.         void m_func1() 
  7.         { 
  8.             cout << "func1n";
  9.         }
  10.         int m_num;
  11.     };
  12.     void (Testpm :: *pmfn)() = &Testpm :: m_func1;
  13.     int Testpm :: *pmd = &Testpm :: m_num;
  14.     int main() 
  15.     {
  16.         Testpm ATestpm;
  17.         Testpm *pTestpm = new Testpm;
  18.         (ATestpm.*pmfn)();
  19.         (pTestpm ->* pmfn)();
  20.         ATestpm.*pmd = 1;
  21.         pTestpm ->* pmd = 2;
  22.         cout << ATestpm.*pmd << endl
  23.         << pTestpm ->* pmd << endl;
  24.     }

a) func1
b) 

   func1
   func1

c) 

   1
   2

d) 

   func1
   func1
   1
   2

View Answer

Answer: d
Clarification: In this program, As we are passing the value twice to the method
in the class, It is printing the func1 twice and then it is printing the given
value.
Output:

$ g++ ptm1.cpp
$ a.out
func1
func1
1
2

 
 

6. What will be the output of the following C++ code?

  1.     #include 
  2.     using namespace std;
  3.     class Car
  4.     {
  5.         public:
  6.         int speed;
  7.     };
  8.     int main()
  9.     {
  10.         int Car :: *pSpeed = &Car :: speed;
  11.         Car c1;
  12.         c1.speed = 1;           
  13.         cout << c1.speed << endl;
  14.         c1.*pSpeed = 2;     
  15.         cout  << c1.speed << endl;
  16.         return 0;
  17.     }

a) 1
b) 2
c) Both 1 & 2
d) 4
Answer: c
Clarification: In this program, We are printing the value by direct access and another one by using pointer to member.
Output:

$ g++ ptm2.cpp
$ a.out
1
2

7. What will be the output of the following C++ code?

  1.     #include 
  2.     using namespace std;
  3.     class bowl 
  4.     {
  5.         public:
  6.         int apples;
  7.         int oranges;
  8.     };
  9.     int count_fruit(bowl * begin, bowl * end, int bowl :: *fruit)
  10.     {
  11.         int count = 0;
  12.         for (bowl * iterator = begin; iterator != end; ++ iterator)
  13.             count += iterator ->* fruit;
  14.         return count;
  15.     }
  16.     int main()
  17.     {
  18.         bowl bowls[2] = {{ 1, 2 },{ 3, 5 }};
  19.         cout << "I have " << count_fruit(bowls, bowls + 2, & bowl :: apples) << " applesn";
  20.         cout << "I have " << count_fruit(bowls, bowls + 2, & bowl :: oranges) << " orangesn";
  21.         return 0;
  22.     }

a) 

   I have 4 apples
   I have 7 oranges

b) 

   I have 3 apples
   I have 5 oranges

c) 

   I have 1 apples
   I have 5 oranges

d) 

   I have 1 apples
   I have 7 oranges

View Answer

Answer: a
Clarification: In this program, We are passing the value to the class and adding the values and printing it in the main.
Output:

$ g++ ptm3.cpp
$ a.out
I have 4 apples
I have 7 oranges

 
 

8. What will be the output of the following C++ code?

  1.     #include 
  2.     using namespace std;
  3.     class Foo
  4.     {
  5.         public:
  6.         Foo(int i = 0){ _i = i;}
  7.         void f()
  8.         {
  9.             cout << "Executed"<<endl;
  10.         }
  11.         private:
  12.         int _i;
  13.     };
  14.     int main()
  15.     {
  16.         Foo *p = 0;
  17.         p -> f();
  18.     }

a) Executed
b) Error
c) Runtime error
d) 10
Answer: a
Clarification: In this program, We passes the value to the class and printing it.
Output:

$ g++ ptm4.cpp
$ a.out
Executed