Question

我正在学习c ++，这令我困惑。我有一个Vector类，加上了加号和插入运算符：

#include <iostream>

class Vector {
    public:
        Vector(float _x, float _y, float _z) {
            x = _x; y = _y; z = _z;
        }

        float x, y, z;
};

Vector operator+(const Vector &v1, const Vector &v2) {
    return Vector(v1.x + v2.x, v1.y + v2.y, v1.z + v2.z);
}

std::ostream &operator<<(std::ostream &out, Vector &v) {
    out << "(" << v.x << ", " << v.y << ", " << v.z << ")";

    return out;
}


int main() {
    Vector i(1, 0, 0);
    Vector j(0, 1, 0);

    std::cout << i;
    /* std::cout << (i + j); */

}

当我尝试打印Vector时，一切都很好：

Vector i(1, 0, 0);
std::cout << i;  // => "(1, 0, 0)"

添加矢量也很好：

Vector i(1, 0, 0);
Vector j(0, 1, 0);
Vector x = i + j;
std::cout << x;  // => "(1, 1, 0)"

但是如果我尝试在没有中间变量的情况下打印两个向量的总和，我会得到巨大的编译错误，我真的不明白：

Vector i(1, 0, 0);
Vector j(0, 1, 0);
std::cout << (i + j);  // Compile Error

vector.cpp: In function ‘int main()’:
vector.cpp:28:15: error: no match for ‘operator<<’ (operand types are ‘std::ostream {aka std::basic_ostream<char>}’ and ‘Vector’)
     std::cout << (i + j);
               ^
vector.cpp:17:15: note: candidate: std::ostream& operator<<(std::ostream&, Vector&) <near match>
 std::ostream &operator<<(std::ostream &out, Vector &v) {
               ^
vector.cpp:17:15: note:   conversion of argument 2 would be ill-formed:
vector.cpp:28:21: error: invalid initialization of non-const reference of type ‘Vector&’ from an rvalue of type ‘Vector’
     std::cout << (i + j);

我做错了什么？这应该工作吗？

Answer 1

加法运算符的结果不是你可以采用非const引用的。但是，由于您未在Vector中修改<<，因此您可以而且应该将其const：

std::ostream &operator<<(std::ostream &out, const Vector &v) {
    out << "(" << v.x << ", " << v.y << ", " << v.z << ")";

    return out;
}

重载插入（＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆gt;）时出错

1 个答案: