使用重载运算符<< with operator*时出错

Question

我最近尝试过运算符重载,并查看了有关运算符重载的此stackoverflow页面(http://stackoverflow.com/questions/4421706/operator-overloading).

我重载了*运算符,可以运行代码如

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

但得到编译时错误 error: invalid operands to binary expression ('basic_ostream<char, std::char_traits<char> >' and 'Vector2')

运行代码时如

std::cout << a*b;

这是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 << ")";
}

这是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);

我不知道从哪里开始.

Answer 1

问题是

a*b

返回一个临时的,所以你需要:

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

作为临时不能绑定到非const引用.