Doubt with Operator Overload Exercise + and -

2

The exercise asks me to implement the addition and subtraction each with a single member method. And you have to be able to execute the following code:

        int main()
        {
            Punto p(12.34,-56.78);
            Punto r,s;
            s=78+p;
            r=78-p;
            cout<<"1. punto p= ("<<p.getX()<<";"<<p.getY()<<")"<<endl;
            cout<<"2. punto 78+p: s= ("<<s.getX()<<";"<<s.getY()<<")"<<endl;
            cout<<"3. punto 78-p: r= ("<<r.getX()<<";"<<r.getY()<<")"<<endl;
            r=p+s-45;
            cout<<"4. punto p+s-45: r= ("<<r.getX()<<";"<<r.getY()<<")"<<endl;
        }

How can the operator overload be done so that the operation can be done in both ways ( int+Punto and Punto+int ) with a single method?

    
asked by Fokker 15.04.2018 в 21:05
source

1 answer

1
  

The exercise asks me to implement the addition and subtraction of each one with a single member method.

It is not the same int + Punto as Punto + int . These cases are going to be treated by independent functions. In the case of the member function, the operator, expressed explicitly, would remain as follows:

Punto Punto::operator+(this,int);

It's easy to see that this function does not work for the case int + Punto since you can not convert an integer in a point and a point in an integer ... To deal with the case int + Punto you need, as we have said, a friend function:

class Punto
{
  friend Punto operator+(int,Punto const& punto); // funcion amiga

  // ...

  Punto operator+(int valor) const; // funcion miembro
};

Punto operator+(int valor, Punto const& punto)
{
  return punto + valor;
}

Punto Punto::operator+(int valor) const
{ /* ... */ }

Note that the function is not a member (and if you try to make it a member you will have a nice compilation error).

    
answered by 16.04.2018 / 09:36
source