C++ | Function Overloading and Default Arguments | Question 4
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
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