如何在同一个类中调用另一个运算符重载成员函数的运算符重载成员函数（或使用运算符）？

Question

I'm writing a code in c++ for handling Complex numbers. I was also practicing operator overloading too. So I overloaded * (multiplication operator), now I want to use the overloaded operator in my overloaded / (division operator), but when I'm using * its showing error. Here is the code:

#include <iostream>
#include <cmath>

using namespace std;

class Imaginary
{
    public:
    //constructors
    Imaginary(double a,double b):x(a),y(b){}
    Imaginary():x(0.0),y(0.0){}

    //setter methods for x and y
    void Setx(double x) { this->x = x; }
    void Sety(double y) { this->y = y; }

    //getter methods for x and y
    double Getx(){return this->x;}
    double Gety(){return this->y;}

    //overloaded operators
    Imaginary operator+(Imaginary&);
    Imaginary operator-(Imaginary&);
    Imaginary operator*(Imaginary&);
    Imaginary operator~();
    Imaginary operator/(Imaginary&);

    void print();
private:
    double x;
    double y;
};

Imaginary Imaginary::operator+(Imaginary &i){
    Imaginary ti;
    ti.Setx(this->x+i.x);
    ti.Sety(this->y+i.y);

    return ti;
}

Imaginary Imaginary::operator-(Imaginary &i){
    Imaginary ti;
    ti.Setx(this->x-i.x);
    ti.Sety(this->y-i.y);
    return ti;
}

Imaginary Imaginary::operator*(Imaginary &i){
Imaginary ti;
ti.Setx((this->x*i.x) - (this->y*i.y));
ti.Sety((this->y*i.x)+(this->x*i.y));
return ti;
}

Imaginary Imaginary::operator~(){
int y;
y = this->y;
this->y = -y;
return *this;
}

Imaginary Imaginary ::operator/(Imaginary &i){
Imaginary numerator,denominator,ti;
//i want to use here the overloaded *(multiplacation) operator
numerator = (*this) * (~i);//showing error
denominator = (*this) * (~i);//showing error
ti.Setx(numerator.Getx()/denominator.Getx());
ti.Sety(numerator.Gety()/denominator.Getx());

return ti;
}

void Imaginary::print(){
cout<<x;
if (y>0)
cout<<"+i"<<y<<endl;
else if (y<0)
cout<<"-i"<<abs(y)<<endl;

}

int main()
{   
Imaginary res;
Imaginary z1(2,3);
Imaginary z2(1,-1);
z1.print();
z2.print();

/*res = z1+z2;
cout<<"Addition:-\n";
res.print();

res = z1-z2;
cout<<"Subtraction:-\n";
res.print()*/

res = z1*z2;
cout<<"Multiplication:-\n";
res.print();

res = z1/z2;
cout<<"Division:-\n";
res.print();

return 0;
}

the error message is following:-

D:\Games\Cheese\main.cpp|66|error: no match for 'operator*' (operand types are 'Imaginary' and 'Imaginary')|

Please anybody show me how to rectify it.

Answer 1

The error isn't the best one in this case. Your operator* is defined as

Imaginary Imaginary::operator*(Imaginary &i)

Which requires the right hand side to be an lvalue. When you do

(*this) * (~i)

in operator/ , (~i) returns Imaginary which is a rvalue. You cannot bind that rvalue to the lavue reference so your overload is not considered and you get a compiler error.

The simplest way to fix this is to take a const & instead like

Imaginary Imaginary::operator*(const Imaginary &i)

Answer 2

Two things

Firstly,

 (*this)*(~i);

is an expression in which the right hand side (result of operator~() ) is a temporary. Your operator*() can only accept a temporary as a reference argument ( Imaginary & )if that argument is declared const .

Also both sides of an operator*() are better specified as const , since - generally speaking - multiplying two values does not change either one (and produces a distinct result). Same goes for other operators (like operator/() ).

The solution is to change the specification of operator*() (member function) to be

Imaginary Imaginary::operator*(const Imaginary &i) const;

Having done that, an alternative way of calling the operator*() would be to replace

numerator = (*this) * (~i);

with

numerator = this->operator*(~i);

or even (since this is happening in a member function) with

numerator = operator*(~i);     //  this-> is implied