Operator overloading+ add two objects

Question

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;

Whe the operator+ is wrong ? BTW: you forget to initialize the members. — Zang MingJie
– Zang MingJie, Commented Jun 27, 2012 at 15:26
You cant do b.num1, num1 is a private member. If you need to access it you need to have a getter function for that. b-> is invalid as you have not created a pointer type. Do you intended to add the private integers? — another.anon.coward
– another.anon.coward, Commented Jun 27, 2012 at 15:27
@another.anon.coward how can I get that? I just want to find a way to do a copy of two objects and then return them in object's name r — Ali
– Ali, Commented Jun 27, 2012 at 15:28
I am not sorry, I dont quite understand. What should operator + add? To get private create a public function which returns the private member maybe - public: inline int getInt1(){return int1;} — another.anon.coward
– another.anon.coward, Commented Jun 27, 2012 at 15:30
@another.anon.coward in the operator+ it suppose to add two objects together and assign it to object call r then return that to the main function in main the calling of this operator is like this Result = Left + right; — Ali
– Ali, Commented Jun 27, 2012 at 15:32

Robᵩ · Accepted Answer · 2012-06-27 15:49:01Z

6

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";
}

edited Jun 27, 2012 at 15:49

answered Jun 27, 2012 at 15:34

Robᵩ

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9 Comments

Zang MingJie Over a year ago

I don't think you should modify the lhs

Robᵩ Over a year ago

I don't. In both instances I modify a local copy of the lhs. In the first instance, I make the copy explicitly. In the second instance, I ask the compiler to make a copy (because lhs is passed by value, not passed by reference.)

Ali Over a year ago

It works, but my result still wrong, probably my another function is not complete.

Robᵩ Over a year ago

The operator I list depends upon a working copy constructor and a working operator +=. Yes, if either of those are incorrect, the result will be incorrect.

Robᵩ Over a year ago

@Ali - I've included a sample program that might help you.

|

Zang MingJie · Accepted Answer · 2012-06-27 15:34:18Z

1

I can't see why the operator+ is wrong:

#include <stdio.h>

class One {
    public:
        One(int n1, int n2): num1(n1),num2(n2) {}

    private:
        int num1, num2;
    public:
        One operator+=(const One& o) {
            num1 += o.num1;
            num2 += o.num2;
            return *this;
        }
        friend bool operator==(const One&, const One&); // - a friend operator that compares two One class objects for equality
        void print() {
            printf("%d,%d\n", num1, num2);
        }
};

One operator+(const One& a, const One& b) {
    One r(0,0);
    r += b;
    r += a;
    return r;
}

int main() {
    One a(1,2),b(3,4);
    One r = a + b;
    r.print();
}

answered Jun 27, 2012 at 15:34

Zang MingJie

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2 Comments

Mike Seymour Over a year ago

It's usual to copy one argument and then add the other to it (i.e. One r=a; r+=b;). That will work even if the type doesn't have an additive identity.

another.anon.coward Over a year ago

#include <cstdio> maybe? :)

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