Peronal Link: http://segmentfault.com/a/1190000002464822
这节课讲了本门课程 面向对象程序设计中最为重要的一个部分 - 多态
1 /*************************************************************************
2 > File Name: polymorphism.cpp
3 > Author: Jeremy Wu
4 > Created Time: Mon 25 May 2015 10:04:07 AM CST
5 > Function: tune every instrument
6 ************************************************************************/
7
8 //多态：一个接口，多种实现
9 //发生多态现象的条件：
10 // １．通过调用虚函数(基类中必须要有，派生类中可有可无建议加上)
11 // ２．借助指针或者引用调用该虚函数
12 //
13 //虚函数只能是 非静态 成员函数
14 //如果类中有虚函数，最好把 析构函数 也定义为虚函数
15 //
16 //如果基类的析构函数是虚函数，由它派生的所有的派生类的析构函数都 自动 成为虚函数
17 // ·构造函数无法定义为虚函数
18 //
19 //基类可以指向它的所有的派生类
20 //相当于基类给所有的派生类提供了统一的接口
21 //一般不会直接创建基类的对象
22 //
23 //{ 抽象基类 }
24 //如果一个类中包含纯虚函数，就是抽象类
25 //抽象类无法创建对象
26 //如果需要 避免 创建基类对象，应该使用纯虚函数
27
28
29
30 //接下来是一个例子，有一个基类叫做乐器类
31 //由它派生出Wind ， Brass 和 Gitar 三个类
32
33 #include <iostream>
34
35 using namespace std;
36
37 class Instrument {
38 public:
39 virtual void play (); //virtual 关键字
40 Instrument () { cout << "Instement () " << endl; }
41 virtual ~Instrument () { cout << "~Instement () " << endl; }
42 };
43
44 void Instrument::play () { //类的外部定义前不用加 virtual 关键字
45 cout << "Instrument::play ()" << endl;
46 }
47
48 class Wind : public Instrument{
49 virtual void play () { cout << "Wind::play () " << endl; } //派生类中 virtual 关键字可加可不加
50 ~Wind () { cout << "Wind::~Wind ()" << endl; }
51 };
52
53 class Brass : public Instrument {
54 virtual void play () { cout << "Brass::play ()" << endl; }
55 ~Brass () { cout << "Brass::~Brass () " << endl; }
56 };
57
58 class Gitar : public Instrument {
59 virtual void play () { cout << "Gitar::play ()" << endl; }
60 ~Gitar () { cout << "Gitar::~Gitar ()" << endl; }
61 };
62
63 //inst 指针在定义的时候，是静态类型
64 //在程序运行过程中，它的动态类型就会根据指向对象的类型作改变
65 //
66 //有时候把动态绑定称为运行时绑定，也称后继绑定
67 //run-time binding
68
69 void tune (Instrument * inst) { //引用传递 或者用指针
70 inst->play (); //Dynamic binding
71 //在编译时进行 动态绑定
72 }
73
74 int main () {
75
76 Instrument * p = new Gitar;
77 Instrument * p2 = new Wind;
78
79 delete p;
80 delete p2;
81
82 return 0;
83 }
接下来老师出了一道题目
描述如下：
以下是我的 Solution:
1 /*************************************************************************
2 > File Name: ployPROB.cpp
3 > Author: Jeremy Wu
4 > Created Time: Mon 25 May 2015 10:50:07 AM CST
5 ************************************************************************/
6
7 //定义一个形状类的继承结构
8 //
9 //Shape -> Triangle
10 // -> Rectangle
11 // -> Circle
12 //
13 //给形状定义两个操作：
14 // getArea () 求面积
15 // getLength () 求周长
16
17 #include <iostream>
18 #include <cmath>
19
20 const double PI = 3.1415926;
21
22 using namespace std;
23
24 class Shape {
25 public:
26 virtual double getArea () = 0;
27 virtual double getLength () = 0;
28 virtual ~Shape () { cout << "~Shape ()" << endl; }
29 Shape () { cout << "Shape ()" << endl; }
30 };
31
32 class Triangle : public Shape {
33 private:
34 double x, y, z;
35 public:
36 double getArea () {
37 double p = (x + y + z) / 2.0;
38 return sqrt (p * (p - x) * (p - y) * (p - z));
39 }
40 double getLength () { return x + y + z; }
41 Triangle (double ix = 0.0, double iy = 0.0, double iz = 0.0) {
42 x = ix, y = iy, z = iz;
43 }
44 ~Triangle () { cout << "~Triangle () " << endl; }
45 };
46
47 class Rectangle : public Shape {
48 private:
49 double x, y;
50 public:
51 double getArea () { return x * y; }
52 double getLength () { return 2.0 * (x + y); }
53 Rectangle (double ix = 0.0, double iy = 0.0) {
54 x = ix, y = iy;
55 }
56 ~Rectangle () { cout << "~Rectangle () " << endl; }
57 };
58
59 class Circle : public Shape {
60 private:
61 double r;
62 public:
63 double getArea () { return PI * r * r; }
64 double getLength () { return 2.0 * PI * r; }
65 Circle (double ir = 0.0) {
66 r = ir;
67 }
68 ~Circle () { cout << "~Circle ()" << endl; }
69 };
70
71 void test (Shape & sp) {
72 cout << "/***********************************/" << endl;
73 cout << ‘\t‘ << "Area is " << sp.getArea () << endl;
74 cout << ‘\t‘ << "Length is " << sp.getLength () << endl;
75 cout << "/***********************************/" << endl;
76 }
77
78 int main () {
79
80 Circle p1 (3.0);
81 Rectangle p2 (3.0, 6.0);
82 Triangle p3 (1.0, 2.0, 1.5);
83
84 test (p1);
85 test (p2);
86 test (p3);
87
88 return 0;
89 }
以下是官方版 Solution:
1 /*************************************************************************
2 > File Name: example.cpp
3 > Author: Jeremy Wu
4 > Created Time: Mon 25 May 2015 11:34:47 AM CST
5 ************************************************************************/
6
7 #include <iostream>
8 #include <cmath>
9
10 using namespace std;
11
12 class Shape {
13 public:
14 virtual double area () = 0; //不实现基类可以这么写，表示定义一个纯虚函数
15 virtual double zhou () = 0; //pure virtual function
16 };
17
18 class Triangle : public Shape {
19 double a, b, c;
20 public:
21 Triangle (double aa, double bb, double cc) : a (aa), b (bb), c (cc) {}
22 virtual double area () {
23 double p = 0.5 * (a + b + c);
24 return sqrt (p * (p - a) * (p - b) * (p - c));
25 }
26 virtual double zhou () {
27 return a + b + c;
28 }
29 };
30
31 class Rectangle : public Shape {
32 double a, b;
33 public:
34 Rectangle (double aa, double bb) : a (aa), b (bb) {}
35 virtual double area () {
36 return a * b;
37 }
38 virtual double zhou () {
39 return 2.0 * (a + b);
40 }
41 };
42
43 class Circle : public Shape {
44 double r;
45 public:
46 Circle (double rr) : r (rr) {}
47 virtual double area () {
48 return 3.14159 * r * r;
49 }
50 virtual double zhou () {
51 return 2.0 * 3.14159 * r;
52 }
53 };
54
55 void printArea (Shape * p) {
56 cout << p->area () << endl;
57 }
58
59 void printZhou (Shape * p) {
60 cout << p->zhou () << endl;
61 }
62
63 int main () {
64
65 Shape * p1, * p2, * p3;
66
67 //Shape sp;
68
69 p1 = new Triangle (6, 7, 8);
70 p2 = new Rectangle (10, 4.6);
71 p3 = new Circle (5.8);
72
73 printArea (p1);
74 printArea (p2);
75 printArea (p3);
76
77 printZhou (p1);
78 printZhou (p2);
79 printZhou (p3);
80
81 return 0;
82 }
时间： 2024-10-07 13:22:57