程序設計教程 第2版教學素材講稿教學 繼承和派生例子

1、例12.1 先定義“點”類Point，再由“點”類派生出“圓”類Circle。#include <iostream.h>#define PI 3.14159class Point / 定義“點”類int x, y;public:Point(int a=0, int b=0)x=a; y=b;void ShowPoint( )cout<<"Point:("<<x<<','<<y<<")n"int Getx( ) return x; int Gety( )return y

2、;void Setxy(int a, int b)x=a; y=b;class Circle: public Point / 定義“圓”類，公有繼承int r; / “圓”的半徑 public:Circle(int x, int y, int ra) : Point(x, y) / B r = ra; void Setr(int ra)r = ra; double Area( ) /求圓的面積return PI*r*r;void Move(int x_offset, int y_offset) /將圓心坐標平移int x1=Getx( ); /存取基類的私有成員int y1=Gety( );

3、/ Dx1 += x_offset; y1 += y_offset;Setxy(x1, y1); / Evoid ShowCircle( )ShowPoint( ); / Fcout<<" Radius: "<<r<<'t'cout<<"Area: "<<Area( )<<endl; /G;void main( )Circle c(1, 1, 1);c.ShowCircle( );c.Move(1, 2);c.ShowCircle( );c.Setxy(4, 5);

4、/重新置圓心坐標 c.Setr(2); /重新置半徑值 c.ShowCircle( );回ppt講稿 例12.2 先定義“點”類Point和“半徑”類Radius，再由Point類和Radius類多重派生出“圓”類Circle。#include <iostream.h>#define PI 3.14159class Point protected: /A int x, y;public:Point(int a=0, int b=0) x=a; y=b; void ShowPoint( ) cout<<"Point:("<<x<<

5、;','<<y<<")n" int Getx( ) return x; int Gety( ) return y; void Setxy(int a, int b) x=a; y=b; ;class Radius protected: /Bint r;public:Radius(int ra=0) r = ra; void Setr(int ra) r = ra; int Getr( ) return r; ;class Circle : public Point, public Radius public:Circle(int x,

6、 int y, int ra) : Point(x, y), Radius(ra) /D double Area( ) return PI*r*r; /直接訪問基類的保護成員 void Move(int x_offset, int y_offset) x += x_offset; y += y_offset; void ShowCircle( ) ShowPoint( );cout<<"Radius: "<<r<<'t'cout<<"Area: "<<Area( )<<

7、;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( );程序的運行結果為：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講稿 例12.3 多重繼承中基類構造函數(shù)和析構函數(shù)的調用順序#include <iostream.h>cla

8、ss Base1protected: int data1;public:Base1(int a=0)data1 = a; cout<<"Base Constructor1n"Base1( )cout<<"Base Destructor1n" ;class Base2protected: int data2;public:Base2(int a=0)data2 = a; cout<<"Base Constructor2n"Base2( )cout<<"Base Destructo

9、r2n" ;class Derived: public Base1, public Base2 /Aint d;public:Derived(int x, int y, int z):Base1(x), Base2(y) /B d=z; cout<<"Derived Constructorn"Derived( ) cout<<"Derived Destructorn" void Show( ) cout<<data1<<','<<data2<<',&

10、#39;<<d<<endl; ;void main( ) Derived c(1, 2, 3);c.Show( );程序的運行結果是：Base Constructor1Base Constructor2Derived Constructor1, 2, 3Derived DestructorBase Destructor2Base Destructor1返回ppt講稿 例12.4 對象成員構造函數(shù)和析構函數(shù)的調用順序#include <iostream.h>class Base1protected: int data1; public:Base1(int a=

11、8)data1 = a; cout<<data1<<", Base Constructor1n"Base1( )cout<<data1<<", Base Destructor1n"class Base2protected: int data2;public:Base2(int a=9)data2 = a; cout<<data2<<", Base Constructor2n"Base2( )cout<<data2<<", Base

12、 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; cout<<"Derived Constructorn"Derived( ) cout<<"Derived Destructorn"void Show( )cout<<data1<&

13、lt;','<<data2<<','<<d<<endl;void main( )Derived c(1, 2, 3);c.Show( );程序的運行結果是：1, Base Constructor12, Base Constructor23, Base Constructor1 /構造對象成員b1時的輸出 4, Base Constructor1 /構造對象成員b2時的輸出 Derived Constructor1, 2, 3Derived Destructor4, Base Destructor1 /析構對象成員b

14、2時的輸出 3, Base Destructor1 /析構對象成員b1時的輸出 2, Base Destructor21, Base Destructor1返回ppt講稿 例12.7 多層繼承中的沖突，注意本例的繼承關系如右下圖所示：C y, SetAx( ),SetBx( ),Sety( ),Gety( ) A x, Show( ) B x, Show( ) D z, Setz( ),Getz( ) #include <iostream.h>class Aprotected:int x;public:void Show( ) cout << "x="

15、; << x << 'n' ; ;class Bprotected:int x;public:void Show( ) cout << "x=" << x << 'n' ; ;class C: public A, public B /公有繼承 A、B 類 int y;public:void SetAx(int a) A:x=a; void SetBx(int a) B:x=a; void Sety(int b) y=b; int Gety( ) return y; ;class D:

16、 public C /公有繼承 C 類 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);cout<<"A" d.C:A:Show( ); /E 報錯 cout<<"B" d.C:B:Show( ); /F 報錯 cout << "y=" << d.Gety( ) << '

17、;n'cout << "z=" << d.Getz( ) << 'n'解決1：在C類中增加成員函數(shù)：void ShowA( ) cout << "x=" << A:x << 'n' void ShowB( ) cout << "x=" << B:x << 'n' 再將E行和F行改寫成：d.ShowA( ); 和d.ShowB( ); 即可。解決2：把E行和F行改寫成：d.A

18、:Show( ); 和d.B:Show( ); 即可。返回ppt講稿 例12.8 支配規(guī)則示例 #include <iostream.h> class Aprotected: int x;public:void Set(int a) x=a; void Show( ) cout << "x=" << x << 'n' ; ;class B : public Aprotected: 問題：類B類對象有幾個數(shù)據(jù)成員。 int x;public:void SetAx(int a) A:x = a; /訪問的是基類A的

19、xvoid SetBx(int a) x = a; /訪問的是派生類B的xvoid Show( ) cout<<"x="<< x <<endl; ;void main(void)B b;b.SetAx(1);b.SetBx(2);b.A:Show( ); /訪問的是基類A的Show( )b.Show( ); /訪問的是派生類B的Show( )返回ppt講稿 例12.11 虛基類和非虛基類構造函數(shù)的調用。#include <iostream.h> class Apublic:A( ) cout<<"A&qu

20、ot;A( ) cout<<"A"class Bpublic:B( ) cout<<"B"B( ) cout<<"B"class Cpublic:C( ) cout<<"C"C( ) cout<<"C"class Dpublic:D( ) cout<<"D"D( ) cout<<"D"class E: virtual public B, public A, public D

21、, virtual public C ;int main( )E c;return 0;程序的運行結果是：BCADDACB返回ppt講稿 例12.12 訪問對象成員的成員#include <iostream.h>#include <math.h>class Pointint x, y;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 Gety( ) return y;void Show( )cout<<

22、;"point("<<x<<','<<y<<")n"class Line Point p1, p2; /對象成員public:Line(int x1, int y1, int x2, int y2): p1(x1, y1), p2(x2, y2)/調用對象成員構造函數(shù) double Length( )int x1, y1, x2, y2;x1=p1.Getx( ); y1=p1.Gety( ); /訪問對象成員p1的成員x2=p2.Getx( ); y2=p2.Gety( ); /訪問對象成

23、員p2的成員return sqrt(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);void Show( )p1.Show( ); /訪問對象成員p1的成員p2.Show( ); /訪問對象成員p2的成員cout<<"Length="<<Length( )<<endl;void main( )Line line(0, 0, 1, 1);line.Show( );本程序的運行結果是：point(0, 0)point(1, 1)Length=1.41421注意在本程序中，Point類和Line類不是繼承關系，只是Point類的兩

24、個對象，是Line類的對象成員，訪問對象成員的成員，與訪問一般對象的成員遵循同樣的規(guī)則。返回ppt講稿 例12.13 訪問基類成員#include <iostream.h>#include <math.h>class Pointprotected:int x, y; /定義x、y為保護成員，以使在公有派生類中可直接訪問它們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 Gety( ) return y; void Show( ) cout<<"point("<<x<<','<<y<<")n" ;class Line : public Point /公有繼承 protected : int x1, y1; public:Line(int a, int b,