C++ Design Patterns
C++ Creational Design Patterns

C++ Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to C++ Framework. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - C++ does not supports the following

A - Multilevel inheritance

B - Hierarchical inheritance

C - Hybrid inheritance

D - None of the above.

Answer : D

Explaination

C++ supports all the forms of inheritance.

Q 2 - By default the members of the structure are

A - private

B - protected

C - public

D - Access specifiers not applicable for structures.

Answer : C

Explaination

If no access specifiers are specified for structure variables/functions, then the default is considered as public.

Q 3 - In the following program f() is overloaded.

void f(int x) {

}

void f(signed x) {

}

main() {

}

A - True

B - False

Answer : B

Explaination

No, as both the functions signature is same.

Q 4 - What is the output of the following program?

#include<iostream>

using namespace std;
class Base {
public:
   void f() {
      cout<<"Base\n";
   }
};

class Derived:public Base {
public:
   void f() {
      cout<<"Derived\n";
   }
};
main() {
   Base *p = new Derived(); 
   
   p->f();
}

A - Base

B - Derived

C - Compile error

D - None of the above.

Answer : A

Explaination

The method f() is not overridden therefore as per the pointer type respective method is called.

#include<iostream>

using namespace std;
class Base {
public:
   void f() {
      cout<<"Base\n";
   }
};
class Derived:public Base {
public:
   void f() {
      cout<<"Derived\n";
   }
};
main() {
   Base *p = new Derived(); 
   
   p->f();
}

Q 5 - What is the output of the following program?

#include<iostream>

using namespace std;
class Base {
public:
   void f() { 
      cout<<"Base\n";
   }
};
class Derived:public Base {
public: 
   void f() {
      cout<<"Derived\n";
   };
};
main() { 
   Derived obj; 
   obj.Base::f();
}

A - Base

B - Derived

C - Compile error

D - None of the above.

Answer : A

Explaination

Base object cannot refer to Derived members.

#include<iostream>

using namespace std;
class Base {
public:
   void f() { 
      cout<<"Base\n";
  }
};
class Derived:public Base {
public: 
   void f() {
      cout<<"Derived\n";
   };
};
main() { 
   Derived obj; 
   obj.Base::f();
}

Q 6 - (i) ios is the base class of istream

(ii) All the files are classified into only 2 types. (1) Text Files (2) Binary Files.

A - Only (i) is true

B - Only (ii) is true

C - Both (i) & (ii) are true

D - Both (i) & (ii) are false

Answer : C

Explaination

Q 7 - What is the output of the following program?

#include <iostream>
using namespace std;
 
int main () {
   // local variable declaration:
   int x = 1;

   switch(x) {
   case 1 :
      cout << "Hi!" << endl; 
      break;
   default :
      cout << "Hello!" << endl;
   }
}

Answer : B

Explaination

Hi, control reaches default-case after comparing the rest of case constants.

#include <iostream>
using namespace std;
 
int main () {
   // local variable declaration:
   int x = 1;

   switch(x) {
   case 1 :
      cout << "Hi!" << endl; 
      break;
   default :
      cout << "Hello!" << endl;
   }
}

Q 8 - What is the output of the following program?

#include<iostream>

using namespace std;
main() { 
   int a[] = {1, 2}, *p = a;
   
   cout<<p[1]; 
}

Answer : B

Explaination

as p holds the base address then we can access array using p just like with a

#include<iostream>

using namespace std;
main() { 
   int a[] = {1, 2}, *p = a;
   
   cout<<p[1]; 
}

Q 9 - The default executable generation on UNIX for a C++ program is ___

Answer : C

Explaination

a.out is the default name of the executable generated on both the UNIX and Linux operating systems.

Q 10 - What is the output of the following program?

#include<iostream>

using namespace std;
main() {
   char s[] = "Fine";
	*s = 'N';
   
   cout<<s<<endl;
}

Answer : B

Explaination

*s=N, changes the character at base address to N.

#include<iostream>

using namespace std;
main() {
   char s[] = "Fine";
	*s = 'N';
   
   cout<<s<<endl;
}

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