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C++ | Function Overloading and Default Arguments | Question 4
  • Last Updated : 03 Aug, 2013

Predict the output of following C++ program.




include<iostream>
using namespace std;
   
class Test
{
protected:
    int x;
public:
    Test (int i):x(i) { }
    void fun() const  { cout << "fun() const " << endl; }
    void fun()        {  cout << "fun() " << endl;     }
};
   
int main()
{
    Test t1 (10);
    const Test t2 (20);
    t1.fun();
    t2.fun();
    return 0;
}


(A) Compiler Error
(B) fun()
fun() const

(C) fun() const
fun() const

(D) fun()
fun()



Answer: (B)

Explanation: The two methods ‘void fun() const’ and ‘void fun()’ have same signature except that one is const and other is not. Also, if we take a closer look at the output, we observe that, ‘const void fun()’ is called on const object and ‘void fun()’ is called on non-const object.
C++ allows member methods to be overloaded on the basis of const type. Overloading on the basis of const type can be useful when a function return reference or pointer. We can make one function const, that returns a const reference or const pointer, other non-const function, that returns non-const reference or pointer. See following for more details.

Function overloading and const keyword

Quiz of this Question

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