CODEWITHSUNDEEP
c++operator-overloading
Evgenij Reznik
Evgenij Reznik

Reputation: 18594

Overloading "+"-operator

I'm trying to overload the + operator in C++, but get the following error:

operators.cpp: In function ‘int main()’:
operators.cpp: 23:17: error: cannot convert ‘Operators’ to ‘int’ in initialization

This is my code:

#include <iostream>
using namespace std;

class Operators{
    private:
        int num1;

    public:
        Operators(int num1){
            this->num1 = num1;
        }
        Operators operator+(Operators o){
            return Operators(num1 + o.num1);
        }
};

int main(){
    Operators o1(5);
    Operators o2(10);
    Operators res = o1 + o2; // EDITED

    cout << res;
}

Could you please help me?

I know, in this case it doesn't make sense to overload it, as I could just say 5+10, but I'm just experimenting.

UPDATE Thanks, I've edited the int.
But now I'm getting the following error:

operators.cpp: In function ‘int main()’:
operators.cpp: 25:10: error: match for ‘operator<<’ in ‘std::cout << res’
[...]

Upvotes: 0

Views: 152

Answers (5)

Andy Prowl
Andy Prowl

Reputation: 126432

The problem with this line:

int res = o1 + o2;

Is that your overload of operator + returns an object of type Operators. This means that you are trying to initialize an int (res) from a temporary of type Operators (the result of o1 + o2), but there is no user-defined conversion for doing that.

This is why the compiler is issuing an error. You can fix this easily by adding a conversion operator to your Operators class:

operator int () const { return num1; }

UPDATE:

It seems you have updated your question. The second line below is problematic:

Operators res = o1 + o2;
cout << res;

Unless you defined an overload of operator << for your Operators class, the compiler won't know which overload of operator << to pick for streaming your object of type Operators.

To solve the issue, you can:

  • Define a conversion operator to int, as suggested above

  • Define an overload of operator << for Operators, as follows:

    friend std::ostream& operator << (std::ostream& o, Operators const& op)
{
    o << op.num1;
    return o;
}

Here is a live example.

Upvotes: 4

David G
David G

Reputation: 96810

Change int to Operators, and implement your own copy and default constructor.

#include <iostream>

class Operators
{
    private:
        int num1;

    public:
        Operators() = default;
        Operators(Operators const& op) : num1(op.num1) {}
        Operators(int num1)
        {
            this->num1 = num1;
        }
        Operators operator+(Operators o)
        {
            return num1 + o.num1;
        }
};

int main()
{
    Operators o1(5);
    Operators o2(10), res;

    res = o1 + o2;

    std::cout << res;
}

To make this work further, create an overload of operator<< and print the sum.

Upvotes: 0

k1ko
k1ko

Reputation: 1

You need to write:

int res = o1.getMethod() + o2.getMethod();

where getMethod() is a public method in your class that returns the private integer num1. Otherwise, you don't have access to it. Try it.

Upvotes: 0

Joseph Mansfield
Joseph Mansfield

Reputation: 110658

Your operator+ returns an Operators object. You then try and assign this result to int res. YOu have given no way to convert from an Operators object to an int. You could provide a conversion operator to do this:

operator int() {
  return num1;
}

This defines a conversion from Operators to int.

Upvotes: 1

SLaks
SLaks

Reputation: 887453

As the error message is trying to tell you, o1 + o2 is of type Operators.
You can't assign that to an int

Upvotes: 3

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