Question

我正在使用带有类的数据模型：Point2D，Point3D，PointGeo：

template <class T>
class Point2D
{
    protected:
            T x;
            T y;
...
};


template <class T>
class Point3D
{
    protected:
            T x;
            T y;
            T z;
...
};

template <class T>
class PointGeo
{
    protected:
            T lat;
            T lon;
...
};

要管理这些类的实例，可以使用允许从文件加载点，添加/删除点，清除列表，打印的下列类...

2D点列表

template <class T>
struct TPoints2DList
{
    typedef std::vector <Point2D <T> > Type;
};


template <class T>
class Points2DList
{
    private:
            typename TPoints2DList <T>::Type  points;


    public:
            Points2DList() : points ( 0 ) {}
            virtual ~Points2DList() {points.clear();}
            Points2DList ( const Points2DList &source );
            typename TPoints2DList <T>::Type ::iterator begin() { return points.begin(); }
            typename TPoints2DList <T>::Type::const_iterator begin() const { return points.begin(); }
            typename TPoints2DList <T>::Type::iterator end() { return points.end(); }
            typename TPoints2DList <T>::Type::const_iterator end() const { return points.end(); }
            Point2D <T> &operator [] ( int index ) {return points[index];}
            const Point2D <T> &operator [] ( int index ) const {return points[index];}

    public:
            //Overloaded member functions
            inline void clear() {points.clear();};
            inline void pop_back() {points.pop_back();}
            inline void push_back ( Point2D <T> p ) { points.push_back ( p );}
            inline unsigned int size() const {return points.size();}

    public:
            //Other methods
            void loadPointsFromFile ( const char *file);
...
}

3D点列表

template <class T>
struct TPoints3DList
{
    typedef std::vector <Point3D <T> > Type;
};


template <class T>
class Points3DList
{
    private:
            typename TPoints3DList <T>::Type  points;


    public:
            Points3DList() : points ( 0 ) {}
            virtual ~Points2DList() {points.clear();}
            Points3DList ( const Points3DList &source );
            typename TPoints3DList <T>::Type ::iterator begin() { return points.begin(); }
            typename TPoints3DList <T>::Type::const_iterator begin() const { return points.begin(); }
            typename TPoints3DList <T>::Type::iterator end() { return points.end(); }
            typename TPoints3DList <T>::Type::const_iterator end() const { return points.end(); }
            Point3D <T> &operator [] ( int index ) {return points[index];}
            const Point3D <T> &operator [] ( int index ) const {return points[index];}

    public:
            inline void clear() {points.clear();};
            inline void pop_back() {points.pop_back();}
            inline void push_back ( Point3D <T> p ) { points.push_back ( p );}
            inline unsigned int size() const {return points.size();}

    public:
            //Other methods
            void loadPointsFromFile ( const char *file);
 ...
}

PointGeo类的源代码类似......

因此，类之间的代码差异很小。它们在加载，打印和保存数据方面有所不同。

设计一个替换所有三个类的类是不合适的？如何创建加载方法，打印特定于数据类型的数据？

动态配置也出现类似的情况：Node2D，Node3D，...类。类Node2D存储一些拓扑关系并使用指向其他节点或面的指针......在这种情况下，所有三个类都将具有不同的析构函数......

2D点列表

template <class T>
struct TNodes2DList
{
    typedef std::vector <Node2D <T> *> Type;
};

非常感谢您的意见和建议。我正在编写几何库并思考最合适的数据模型。

Answer 1

您可以将I / O操作直接置于Point2D Point3D类型。然后你就不必创建额外的列表类，因为打印/阅读就像这样简单：

for (std::vector<Point3D>::iterator i = a.begin; i != a.end(); ++i) {
    i->print_to_file(file);
}

如果这不可行，至少可以使用相同的列表类作为模板来同时为Point3D和Point2D提供

Answer 2

根据您对课程的处理方式，您可以执行以下操作：

template<class T, unsigned int count = 2>
class Point
{
public:
    T t[count];
    // other data members here
};

然而，如果你正在做类似矢量数学的事情（例如交叉产品，点积，规范化等），这样做会使类比具有2个类（2D和3D）复杂得多。除非你在Geo类中做一些专门的工作，否则根本不需要有不同的2D和Geo形式。

C ++，数据模型，模板

2 个答案: