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.

Quiz of this Question

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