C++ | Operator Overloading | Question 6
Predict the output
#include<iostream>using namespace std;class A{ int i;public: A(int ii = 0) : i(ii) {} void show() { cout << i << endl; }}; class B{ int x;public: B(int xx) : x(xx) {} operator A() const { return A(x); }}; void g(A a){ a.show();} int main(){ B b(10); g(b); g(20); return 0;} |
(A) Compiler Error
(B)
10 20
(C)
20 20
(D)
10 10
Answer: (B)
Explanation: Note that the class B has as conversion operator overloaded, so an object of B can be converted to that of A.
Also, class A has a constructor which can be called with single integer argument, so an int can be converted to A.
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