c++ - 允许继承类使用相同的函数

Question

我有一个由 EulerAngle 类继承的 Vector 类。我希望能够在 EulerAngle 对象上使用 Vector 类中的运算符函数，而不必重写 EulerAngle 类中的运算符并尽可能将 Vector3 更改为 Angle。我怎么能这样做？

class Vector3 {
public:  
    float x, y, z;

    // ctors...
    __forceinline Vector3() : x{}, y{}, z{} {

    __forceinline Vector3( float x, float y, float z ) : x{ x }, y{ y }, z{ z } {

    }

    // functions...

    // operators...
    __forceinline Vector3 &operator +=( const Vector3 &other ) {
        x += other.x;
        y += other.y;
        z += other.z;

        return *this;
    }

    __forceinline Vector3 &operator +=( float scalar ) {
        x += scalar;
        y += scalar;
        z += scalar;

        return *this;
    }

    __forceinline Vector3 operator +( float scalar ) const {
       return Vector3( x + scalar, y + scalar, z + scalar );
    }

    __forceinline Vector3 operator +( const Vector3 &other ) const {
       return Vector3( x * other.x, y * other.y, z * other.z );
    }

}

class EulerAngle : public Vector3 {
private:

};

// example, MSVC says this is an error and that there is no suitable conversion from EulerAngle to Vector3
EulerAngle ang = { 1.f, 1.f, 1.f };
ang += 5.f

在回答了几个问题后，我添加了以下代码来完成这项工作。但是我仍然想知道是否有更好的方法来实现这一点？

class EulerAngle : public Vector3 {
public:
    using Vector3::Vector3;

    __forceinline EulerAngle( const Vector3 &other ) : Vector3( other ) {

    }
};

Answer 1

您可以使用 user-defined conversion从 Vector3 到您想要的类型，当它不是从 Vector3 继承时使用静态断言。

template<typename T>
operator T() const
{
    static_assert(std::is_base_of<Vector3, T>::value, "T is not inherited from Vector3");
    return T{x,y,z};
}

LIVE DEMO

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插件

随着 C++20 和 concepts 的引入，上面的代码可以改写为:

template<typename T>
requires std::DerivedFrom<T,Vector3>
constexpr operator T() const
{
    return T{x,y,z};
}

LIVE DEMO