How Non-Member Functions Improve Encapsulation

Question

I read Scott Meyers\' article on the subject and quite confused about what he is talking about. I have 3 questions here.

Question 1

To expl

Answer 1

The meaning f needs type conversions on it left-most arg is as follows:

consider following senario :

Class Integer 
{  
    private: 
       int num;
     public:
        int getNum( return num;)
        Integer(int n = 0) { num = n;}

    Integer(const Integer &rhs)) { num = rhs.num ;}
    Integer operator * (const Integer &rhs)
    {
         return Integer(num * rhs.num);
    }
}


int main()
{
    Integer i1(5);

    Integer i3 = i1 *  3; // valid 

    Integer i3 = 3 * i1 ; // error     
}

In above code i3 = i1 * 3 is equivalent to this->operator*(3) which is valid since 3 is implicitly converted to Integer.

Where as in later i3 = 3 * i1 is equivalent to 3.operator*(i1) , as per rule when u overload operator using member function the invoking object must be of the same class. but here its not that.

To make Integer i3 = 3 * i1 work one can define non-member function as follow :

Integer operator * (const Integer &lhs , const Integer &rhs) // non-member function
    {

         return Integer(lhs.getNum() * rhs.getNum());

    }

I think you will get idea from this example.....