Error when using overloaded operator << with operator *

Question

I've recently tried operator overloading, and have looked at this stackoverflow page (http://stackoverflow.com/questions/4421706/operator-overloading) about operator overloading.

I have overloaded the * operator and can run code such as

Vector2 a(2, 3);
Vector2 b(5, 8);
Vector2 c = a*b;

but get the compile time error error: invalid operands to binary expression ('basic_ostream<char, std::char_traits<char> >' and 'Vector2')

When running code such as

std::cout << a*b;

Here is Vector2.cpp

#include "Vector2.h"

Vector2::Vector2(const float x, const float y) {
    this->x = x;
    this->y = y;
}

Vector2 &Vector2::operator*=(const Vector2 &rhs) {
    this->x *= rhs.x;
    this->y *= rhs.y;
    return *this;
}

std::ostream &operator<< (std::ostream &out, Vector2 &vector) {
    return out << "(" << vector.x << ", " << vector.y << ")";
}

and here is Vector2.h

#include <iostream>

class Vector2 {
    public:
        float x;
        float y;

        Vector2(const float x, const float y);
        Vector2 &operator*=(const Vector2 &rhs);
};

inline Vector2 operator*(Vector2 lhs, const Vector2 &rhs) {
    lhs *= rhs;
    return lhs;
}

std::ostream &operator<<(std::ostream &out, Vector2 &vector);

I'm not sure where to go from here.

Here's the proof why not: http://en.cppreference.com/w/cpp/language/operator_precedence — chris, May 25 '12 at 19:45
possible duplicate of [Why won't cout << work with overloaded * operator?](http://stackoverflow.com/questions/451983/why-wont-cout-work-with-overloaded-operator) — Bo Persson, May 26 '12 at 04:15

Luchian Grigore · Accepted Answer · 2012-05-25T19:56:31.127

6

The problem is that

a*b

returns a temporary, so you need:

std::ostream &operator<<(std::ostream &out, const Vector2 &vector);
//                                            |
//                                      notice const

as a temporary can't bind to a non-const reference.

edited May 25 '12 at 19:56

answered May 25 '12 at 19:46

Luchian Grigore

253,575
64
457
625

1

To your second point, why? The way he has it is perfect. You've got to make a copy of the object anyway, so why not do it in the parameter list? – Benjamin Lindley May 25 '12 at 19:49
Thank you, that makes perfect sense. – rabbidrabbit May 25 '12 at 19:50
@BenjaminLindley why make a copy to the object? – Luchian Grigore May 25 '12 at 19:52
@LuchianGrigore: Can you write the function without making a copy? – Benjamin Lindley May 25 '12 at 19:52
@LuchianGrigore I don't understand how you could write the function without making a copy. It's much easier to take a copy. – rabbidrabbit May 25 '12 at 19:54
Ah yes you're both right. :) Dismiss that last part. @BenjaminLindley – Luchian Grigore May 25 '12 at 19:56
As a bonus, this form allows you to take advantage of copy elision and move semantics if the first argument happens to be an r-value. For full benefits, you would need to make another version which takes the rhs by value and the lhs by const ref. – Benjamin Lindley May 25 '12 at 20:05

score 0 · Answer 2 · answered May 25 '12 at 19:46

0

The following should work:

Vector2 c = a*b;
std::cout << c;

answered May 25 '12 at 19:46

Hakan Serce

11,198
3
29
48

Yes that does work. What I am trying to do is to do std::cout << a*b; – rabbidrabbit May 25 '12 at 19:48
Then you should update the operator signature to take const argument as others suggested. – Hakan Serce May 25 '12 at 19:49

Error when using overloaded operator << with operator *

2 Answers2