Friend Function In C++ Mcq Exercise 4

1. Which of the following statements, which would be the scope of friend function in C++?

• ✅ Class scope
• ✅ Global scope
• ✅ Protected scope
• ✅ Local

Explanation:

The correct answer is option (b). The friend functions in the c++ will have the global scope since they are not the members of the class.

1. Can a friend function be declared as const?

• ✅ Both
• ✅ None

Explanation:

The correct answer is option (b). The const variable cannot be used when declaring friend functions because the functions are not member functions of a class and do not apply to any instance of the class.

1. What is the output for the below following C++ code?
Example
Copy
#include<iostream>

using namespace std;

class A 

{

    int x;

public:

    A(int val) : x(val) {}

    friend void operator++(A &a);

};

void operator++(A &a) 

{

    a.x++;

    cout << a.x;

}

int main() 

{

    A obj(10);

    ++obj;

    return 0;

}
Try it Yourself

• ✅ Runtime error

Explanation:

The correct answer is option (b). The friend function operator (++) overloads the increment operator to increment the private member x of class A and prints the result, which is 11.

1. What is the output for the below following C++ Code?
Example
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#include<iostream>

using namespace std;

class A 

{

    int x;

public:

    A() : x(5) {}

    friend void increase(A &a);

    void display() const {

        cout << x;

    }

};

void increase(A &a) 

{

    a.x += 10;

}

int main() 

{

    A obj;

    increase(obj);

    obj.display();

    return 0;

}
Try it Yourself

• ✅ Runtime error

Explanation:

The correct answer is option (c). The friend function increase modifies the private member x of class A, and the display member function prints the updated value, which is 15.

1. Whether the below following syntax is the correct way of declaring the friend function for class A?
Example
Copy
class A {

public:

    void friend display();

};
Try it Yourself

• ✅ Correct
• ✅ Incorrect
• ✅ None of the above
• ✅ Both a and b

Explanation:

The correct answer is option (b). In the correct way of declaring friend function, void keyword is not included before friend.

1. Is it possible to call a friend function using object of a class?

• ✅ Both
• ✅ None

Explanation:

The correct answer is option (b). A friend function is not a member of the class, so it cannot be called using an object of the class. It must be called independently.