C ++避免代码重复运算符重载

Question

我的问题如下，假设我有这个 classes:

class ID3OBJ
{
public:
    const double X;
    const double Y;
    const double Z;
    ID3OBJ(double x, double y, double z);
    ID3OBJ();
    const ID3OBJ operator+ (const ID3OBJ param) const;
}
class Vector : public ID3OBJ
    {
    public:
        Vector(double x, double y, double z);
        Vector();
        /*const Vector const operator+ (const Vector param); */
}
class Point : public ID3OBJ
    {
    public:
        Point(double x, double y, double z);
        Point();
}
class Plane : public ID3OBJ
    {
    public:
        Point Origin;
        Vector Direction1;
        Vector Direction2;
        Vector NormalVector;
        Plane(Point Origin, Vector Normalvector);
        Plane(Point Origin, Vector Direction1, Vector Direction2);
        Plane();
}
class D3GON : public ID3OBJ, public Plane
    {
    public:
        std::vector<Point> EdgePoints;
        D3GON(Point P1, Point P2, Point P3);
    };

在我当前的代码中，我必须为每个 class 重新定义运算符重载，我怎样才能避免这种代码重复？

我必须添加转换函数吗？

我使用 const member-values 来拒绝对象创建后的更改。这意味着如果必须更改任何低级对象，则必须将其替换为新对象。请参阅下面我的运算符重载：

// operator overwrites:
    const ID3OBJ ID3OBJ::operator+ (const ID3OBJ param) const { double newX, newY, newZ; newX = X + param.X; newY = Y + param.Y; newZ = Z + param.Z; return ID3OBJ(newX, newY, newZ); }

谢谢:)

Answer 1

Curiously recurring template pattern 是去这里的路。它稍微复杂一些，因为您需要多个派生级别。但这里有一个代码示例：

template <class T>
class ID3OBJ
{
public:
   double X,Y,Z;
   T operator+(const T& obj) const {
       T t;
       t.X = X + obj.X;
       t.Y=Y+obj.Y;
       t.Z=z+obj.Z;
       return t;
   }
};

class Vector : public ID3OBJ<Vector>
{
public:
    // some stuff
};

class Point : public Vector, public ID3OBJ<Point>
{
public:
    // X, Y and Z exist twice, once in Vector, once in ID3OBJ<Point>, so we must disambiguate
    using ID3OBJ<Point>::X;
    using ID3OBJ<Point>::Y;
    using ID3OBJ<Point>::Z;
};

您可以添加 Vector（您将得到 Vector）或仅添加 Point，您将得到 Point。更有趣的是，如果您添加 Point 和 Vector，您会得到 Vector 结果，因为重载运算符 + 解析将 select Vector::operator +

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