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Home/ Questions/Q 8141429
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T12:30:21+00:00 2026-06-06T12:30:21+00:00

I’m trying to add two object that they are in the same class. In

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I’m trying to add two object that they are in the same class.

In the private section of the class I have two int variables

class One {

private:
int num1, num2;
public:

    One operator+=(const One&); // - a member operator that adds another One object - to the current object and returns a copy of the current object
    friend bool operator==(const One&, const One&); // - a friend operator that compares two One class objects for equality
};

 One operator+(const One&, const One&);// - a non-friend helper operator that adds One objects without changing their values and returns a copy of the resulting One 

I’m not sure I have a problem on the opeartor+ I guess

One operator+(const One &a, const One &b){

One c,d,r;

c = a;
d = b;

r += b;
r += a;

return r;
}

I think the above code is wrong, but I tried to use like b.num1 and I get compile error

error: 'int One::num1' is private
error: within this context

and I can’t use b->num1 as well because the above function is not in the member function section.

error: base operand of '->' has non-pointer type 'const One'

This is how it calls in main

Result = LeftObject + RightObject;

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T12:30:23+00:00Added an answer on June 6, 2026 at 12:30 pm

    If you have already implemented this member function:

    One One::operator+=(const One&);
    

    Then you may implement the non-member addition operator thus:

    One operator+(const One& lhs, const One& rhs) {
      One result = lhs;
      result += rhs;
      return result;
    }
    

    This can be simplified somewhat into the following:

    One operator+(One lhs, const One& rhs) {
      return lhs += rhs;
    }
    

    This pattern (which you can adapt for all operator/operator-assignment pairs) declares the operator-assignment version as a member — it can access the private members. It declares the operator version as a non-friend non-member — this allows type promotion on either side of the operator.

    Aside: The += method should return a reference to *this, not a copy. So its declaration should be: One& operator+(const One&).


    EDIT: A working sample program follows.

    #include <iostream>
    class One {
    private:
      int num1, num2;
    
    public:
      One(int num1, int num2) : num1(num1), num2(num2) {}
      One& operator += (const One&);
      friend bool operator==(const One&, const One&);
      friend std::ostream& operator<<(std::ostream&, const One&);
    };
    
    std::ostream&
    operator<<(std::ostream& os, const One& rhs) {
      return os << "(" << rhs.num1 << "@" << rhs.num2 << ")";
    }
    
    One& One::operator+=(const One& rhs) {
      num1 += rhs.num1;
      num2 += rhs.num2;
      return *this;
    }
    
    One operator+(One lhs, const One &rhs)
    {
      return lhs+=rhs;
    }
    
    int main () {
      One x(1,2), z(3,4);
      std::cout << x << " + " << z << " => " << (x+z) << "\n";
    }
    
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