Class 继承和运算符重载

Question

我正在通过 SFML 上的一个小项目学习 C++，我想扩展 sf::Vector2<T> class，它表示一个二维数学向量。这个class的声明是available here。特别是，我想向 class 添加一些方法，用于范数计算、旋转等

这是我目前所拥有的：

#include <cmath>
#include <SFML/System/Vector2.hpp>
#include <SFML/Graphics/Transform.hpp>
#include <ostream>

namespace dw::math {

template <typename T>
class Vector2 : public sf::Vector2<T> {
public:
    // Don't really know what does that exactly means, explainations are welcome !
    using sf::Vector2<T>::Vector2;

    template <typename U>
    explicit Vector2(const U &vector) : sf::Vector2<T>(vector) {}

    // This ctor is for implicit conversion from base, don't know if it's really useful ?
    Vector2(sf::Vector2<T> &vector) : sf::Vector2<T>(vector) {}

    // Some methods here ...

    friend std::ostream & operator<<(std::ostream &os, const math::Vector2<T>& right) {
        os << '{' << right.x << ", " <<right.y << '}';
        return os;
    }
};

稍后在代码中，我声明了这样一个结构：

struct Derivative {
    dw::math::Vector2f dPos;
    dw::math::Vector2f dVel;

    Derivative() = default;
    Derivative(const dw::math::Vector2f &dPos, const dw::math::Vector2f &dVel) : dPos(dPos), dVel(dVel) {}

    Derivative operator+(const Derivative &rhs) const {
        return Derivative(
            dPos + rhs.dPos,
            dVel + rhs.dVel
        );
    }
}

Derivative 的 operator+ 重载的 return 部分无法编译：

/home/palra/Documents/Projets/dw/src/physics/World.cpp: In member function ‘Derivative Derivative::operator+(const Derivative&) const’:
/home/palra/Documents/Projets/dw/src/physics/World.cpp:24:18: error: invalid user-defined conversion from ‘sf::Vector2<float>’ to ‘const Vector2f& {aka const dw::math::Vector2<float>&}’ [-fpermissive]
             dPos + rhs.dPos,
             ~~~~~^~~~~~~~~~
In file included from /home/palra/Documents/Projets/dw/src/physics/Body.h:8:0,
                 from /home/palra/Documents/Projets/dw/src/physics/World.h:10,
                 from /home/palra/Documents/Projets/dw/src/physics/World.cpp:5:
/home/palra/Documents/Projets/dw/src/physics/../math/Vector2.h:30:5: note: candidate is: dw::math::Vector2<T, <template-parameter-1-2> >::Vector2(sf::Vector2<T>&) [with T = float; <template-parameter-1-2> = void] <near match>
     Vector2(sf::Vector2<T> &vector) : sf::Vector2<T>(vector) {}

我不明白为什么这不起作用。表达式 dPos + rhs.dPos 与 sf::Vector2<float> operator +(const sf::Vector2<float>& left, const sf::Vector2<float>& right) 的调用兼容，因为 dw::math::Vector2<float> 是继承的 sf::Vector2<float>。然后这个表达式应该产生一个 sf::Vector2<float>，这要归功于我猜的非显式构造函数，它可以分配给 dw::math::Vector2f。我哪里错了？我该如何让它发挥作用？

Answer 1

sf::Vector<T> 并未设计为派生 class，因此更好的方法是编写扩展 sf::Vector<T> 功能的免费函数，这可以无缝地用于运营商。不幸的是，对于正常的函数调用，C++ 还不支持像 C# 那样的扩展。

Class 继承和运算符重载

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