《C++程序设计教程 第2版》教学素材PPT讲稿教学课件第12章 继承和派生例子.docx

1、C+程序设计教程 第2版教学素材PPT讲稿教学课件第12章 继承和派生例子例12.1 先定义“点”类Point，再由“点”类派生出“圆”类Circle。#include #define PI 3.14159class Point / 定义“点”类 int x, y;public: Point(int a=0, int b=0) x=a; y=b; void ShowPoint( ) coutPoint:(x,y)n; int Getx( ) return x; int Gety( ) return y; void Setxy(int a, int b) x=a; y=b; ;class Cir

2、cle: public Point / 定义“圆”类，公有继承 int r; / “圆”的半径 public: Circle(int x, int y, int ra) : Point(x, y) / B r = ra; void Setr(int ra) r = ra; double Area( ) /求圆的面积 return PI*r*r; void Move(int x_offset, int y_offset) /将圆心坐标平移 int x1=Getx( ); /存取基类的私有成员 int y1=Gety( ); / D x1 += x_offset; y1 += y_offset;

3、Setxy(x1, y1); / E void ShowCircle( ) ShowPoint( ); / F cout Radius: rt; coutArea: Area( )endl; /G ;void main( ) Circle c(1, 1, 1); c.ShowCircle( ); c.Move(1, 2); c.ShowCircle( ); c.Setxy(4, 5); /重新置圆心坐标 c.Setr(2); /重新置半径值 c.ShowCircle( );回ppt讲稿 例12.2 先定义“点”类Point和“半径”类Radius，再由Point类和Radius类多重派生出“圆

4、”类Circle。#include #define PI 3.14159class Point protected: /A int x, y;public: Point(int a=0, int b=0) x=a; y=b; void ShowPoint( ) coutPoint:(x,y)n; int Getx( ) return x; int Gety( ) return y; void Setxy(int a, int b) x=a; y=b; ;class Radius protected: /B int r;public: Radius(int ra=0) r = ra; void

5、Setr(int ra) r = ra; int Getr( ) return r; ;class Circle : public Point, public Radius public: Circle(int x, int y, int ra) : Point(x, y), Radius(ra) /D double Area( ) return PI*r*r; /直接访问基类的保护成员 void Move(int x_offset, int y_offset) x += x_offset; y += y_offset; void ShowCircle( ) ShowPoint( ); cou

6、tRadius: rt; coutArea: Area( )endl; ;void main( ) Circle c(1, 1, 1); c.ShowCircle( ); c.Move(1, 2); c.ShowCircle( ); c.Setxy(4, 5); c.Setr(2); c.ShowCircle( );程序的运行结果为：Point:(1, 1)Radius: 1 Area: 3.14159Point:(2, 3)Radius: 1 Area: 3.14159Point:(4, 5)Radius: 2 Area: 12.5664返回ppt讲稿 例12.3 多重继承中基类构造函数和析

7、构函数的调用顺序#include class Base1protected: int data1;public: Base1(int a=0) data1 = a; coutBase Constructor1n; Base1( ) coutBase Destructor1n; ;class Base2protected: int data2;public: Base2(int a=0) data2 = a; coutBase Constructor2n; Base2( ) coutBase Destructor2n; ;class Derived: public Base1, public B

8、ase2 /A int d;public: Derived(int x, int y, int z):Base1(x), Base2(y) /B d=z; coutDerived Constructorn; Derived( ) coutDerived Destructorn; void Show( ) coutdata1,data2,dendl; ;void main( ) Derived c(1, 2, 3); c.Show( );程序的运行结果是：Base Constructor1Base Constructor2Derived Constructor1, 2, 3Derived Des

9、tructorBase Destructor2Base Destructor1返回ppt讲稿 例12.4 对象成员构造函数和析构函数的调用顺序#include class Base1protected: int data1; public: Base1(int a=8) data1 = a; coutdata1, Base Constructor1n; Base1( ) coutdata1, Base Destructor1n; ;class Base2protected: int data2;public: Base2(int a=9) data2 = a; coutdata2, Base

10、Constructor2n; Base2( ) coutdata2, Base Destructor2n; ;class Derived:public Base1, public Base2 /A int d; Base1 b1, b2; /Bpublic: Derived(int x, int y, int z) : Base1(x), Base2(y), b1(x+y), b2(x+z) /C d=z; coutDerived Constructorn; Derived( ) coutDerived Destructorn; void Show( ) coutdata1,data2,den

11、dl; ;void main( ) Derived c(1, 2, 3); c.Show( );程序的运行结果是：1, Base Constructor12, Base Constructor23, Base Constructor1 /构造对象成员b1时的输出 4, Base Constructor1 /构造对象成员b2时的输出 Derived Constructor1, 2, 3Derived Destructor4, Base Destructor1 /析构对象成员b2时的输出 3, Base Destructor1 /析构对象成员b1时的输出 2, Base Destructor21,

12、 Base Destructor1返回ppt讲稿 例12.7 多层继承中的冲突，注意本例的继承关系如右下图所示：#include class Aprotected: int x;public: void Show( ) cout x= x n ; ;class Bprotected: int x;public: void Show( ) cout x= x n ; ;class C: public A, public B /公有继承 A、B 类 int y;public: void SetAx(int a) A:x=a; void SetBx(int a) B:x=a; void Sety(i

13、nt b) y=b; int Gety( ) return y; ;class D: public C /公有继承 C 类 int z;public: void Setz(int a) z=a; int Getz( ) return z; ;void main(void) D d; d.SetAx(10); d.SetBx(20); d.Sety(30); d.Setz(40); coutA; d.C:A:Show( ); /E 报错 coutB; d.C:B:Show( ); /F 报错 cout y= d.Gety( ) n; cout z= d.Getz( ) n;解决1：在C类中增加成

14、员函数： void ShowA( ) cout x= A:x n; void ShowB( ) cout x= B:x n; 再将E行和F行改写成：d.ShowA( ); 和d.ShowB( ); 即可。解决2：把E行和F行改写成：d.A:Show( ); 和d.B:Show( ); 即可。返回ppt讲稿 例12.8 支配规则示例 #include class Aprotected: int x;public: void Set(int a) x=a; void Show( ) cout x= x n ; ;class B : public Aprotected: 问题：类B类对象有几个数据成

15、员。 int x;public: void SetAx(int a) A:x = a; /访问的是基类A的x void SetBx(int a) x = a; /访问的是派生类B的x void Show( ) coutx= x endl; ;void main(void) B b; b.SetAx(1); b.SetBx(2); b.A:Show( ); /访问的是基类A的Show( ) b.Show( ); /访问的是派生类B的Show( )返回ppt讲稿 例12.11 虚基类和非虚基类构造函数的调用。#include class Apublic: A( ) coutA; A( ) cout

16、A;class Bpublic: B( ) coutB; B( ) coutB;class Cpublic: C( ) coutC; C( ) coutC;class Dpublic: D( ) coutD; D( ) coutD;class E: virtual public B, public A, public D, virtual public C ;int main( ) E c; return 0;程序的运行结果是：BCADDACB返回ppt讲稿 例12.12 访问对象成员的成员#include #include class Point int x, y;public: Point

17、(int a=0, int b=0) x=a; y=b; void Setx(int a)x=a; void Sety(int a)y=a; int Getx( ) return x; int Gety( ) return y; void Show( ) coutpoint(x,y)n; ;class Line Point p1, p2; /对象成员public: Line(int x1, int y1, int x2, int y2): p1(x1, y1), p2(x2, y2)/调用对象成员构造函数 double Length( ) int x1, y1, x2, y2; x1=p1.G

18、etx( ); y1=p1.Gety( ); /访问对象成员p1的成员 x2=p2.Getx( ); y2=p2.Gety( ); /访问对象成员p2的成员 return sqrt(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2); void Show( ) p1.Show( ); /访问对象成员p1的成员 p2.Show( ); /访问对象成员p2的成员 coutLength=Length( )endl; ;void main( ) Line line(0, 0, 1, 1); line.Show( );本程序的运行结果是：point(0, 0)point(1, 1)Length

19、=1.41421注意在本程序中，Point类和Line类不是继承关系，只是Point类的两个对象，是Line类的对象成员，访问对象成员的成员，与访问一般对象的成员遵循同样的规则。返回ppt讲稿 例12.13 访问基类成员#include #include class Pointprotected: int x, y; /定义x、y为保护成员，以使在公有派生类中可直接访问它们public: Point(int a=0, int b=0) x=a; y=b; void Setx(int a)x=a; void Sety(int a)y=a; int Getx( ) return x; int Ge

20、ty( ) return y; void Show( ) coutpoint(x,y)n; ;class Line : public Point /公有继承 protected : int x1, y1; public: Line(int a, int b, int c, int d) : Point(a, b) /调用基类构造函数 x1=c; y1=d; double Length( ) return sqrt(x-x1)*(x-x1)+(y-y1)*(y-y1); / A 直接访问基类保护成员x、y void Show( ) Point:Show( ); /访问基类成员函数Show( )

21、coutpoint(x1,y1 )n; coutLength=Length( )endl; ;void main( ) Line line(0, 0, 1, 1); line.Show( );返回ppt讲稿 例12.14 赋值兼容规则#include class Pointprotected: int x, y; / 保护成员public: Point(int a=0, int b=0) x=a; y=b; void Show( ) coutpoint(x,y)n; ;class Line : public Point / 公有继承protected: int x1, y1;public: Line(int a, int b, int c, int d): Point(a, b) /调用基类构造函数 x1=c; y1=d; ;void main( ) Line line(2, 2, 6, 6); Point p; p = line; / A p.Show( );程序运行结果是： Point(2, 2)返回ppt讲稿