填空题

[说明]
现要编写一个画矩形的程序，目前有两个画图程序：DP1和DP2，DP1用函数draw_a_line(x1, y1, x2, y2)画一条直线，DP2则用drawline(x1, x2, y1, y2)画一条直线。当实例化矩形时，确定使用DP1还是DP2。为了适应变化，包括“不同类型的形状”和“不同类型的画图程序”，将抽象部分与实现部分分离，使它们可以独立地变化。这里，“抽象部分”对应“形状”，“实现部分”对应“画图”，与一般的接口(抽象方法)与具体实现不同。这种应用称为Bridge(桥接)模式。图显示了各个类间的关系。
这样，系统始终只处理3个对象：Shape对象、Drawing对象、DP1或DP2对象。以下是C++语言实现，能够正确编译通过。
[C++代码]
class DP1
public:
static void draw_a_line(double x1, double y1,double x2, double y2)
//省略具体实现

;
class DP2
public:
static void drawline(double x1, double x2, double y1, double y2)
//省略具体实现

;
class Drawing
public:
______ void drawLine(double x1,double y1,double x2,double y2)=0
;
class ViDrawing : public Drawing
public:
void drawLine(double x1, double y1, double x2, double y2)
DP1∷draw_a_line(x1, y1, x2, y2);

;
class V2Drawing : public Drawing
public:
void drawLine(double x1, double y1, double x2, double y2)
______;

;
class Shape
private:
______ _dp;
public:
Shape(Drawing *dp);
virtual void draw() = 0;
void drawLine(double x1, double y1, double x2, double y2);
;
Shape∷Shape(Drawing *dp)

_dp : dp;

void Shape∷drawLine(double x1, double y1, double x2, double y2)
//画一条直线
______;

class Rectangle : public Shape
private:
double _x1, _y1, _x2, _y2;
public:
Rectangle(Drawing *dp, double x1, double y1,
double x2, double y2);
void draw();
;
Rectangle∷Rectangle(Drawing *dp, double x1, double y1, double x2, double y2)
: ______

_x1 = x1; _y1 = y1; _x2 = x2; _y2 = y2

void Rectangle∷draw()

//省略具体实现