面向对象程序设计26125

1、目录一、项目名称（从cc+）：27、设计一个函数求2二、项目名称：(类和对象）37、已知一个时钟类cclock定义如下3三、项目名称（继承与派生）：51、已知类cstring和main()函数53、写一个程序，定义抽象基类cshape，由它派生出5个派生类：ccircle(圆形)，csquare(正方形)，crectangle(矩形)，ctrapezoid(梯形)，ctriangle(三角形)，用一个虚函数分别计算几种图形的面积，并求它们的和。要求用基类指针数组，使它每一个元素指向一个派生类对象。10四、项目名称：（opp中的模板）：133、设计一个函数模板mam(t)求一个数组中最大的元素，

2、并以整数数组和字符数组进行调用，并采用相关数据进行测试。13五、心得体会13一、项目名称（从cc+）：7、设计一个函数求程序如下：#include <stdio.h>int fun(int n,int m)int i,j;float s=1.0;for (i=1;i<=m;i+)s=1.0*s*(n-i+1)/i;return (int)s;void main()int m,n,s;printf("input n m (n>m): ");scanf("%d%d",&n,&m);s=fun(n,m);printf(&

3、quot;s = %dn",s);printf("按任意键继续n");scanf("%c",&s);二、项目名称：(类和对象）7、已知一个时钟类cclock定义如下class cclockprivate:int hour,minute,second;public:.void display()cout<<hour<<':'<<minute<<':'<<second<<endl;完成相关成员函数的定义，达到模拟时钟的目的如下图：时钟cc

4、lock#include <iostream>#include <windows.h>using namespace std;class time              private: int h; int m; int s;public: time(); time(int h, int m, int s); time()  cout<<"des

5、truct"<<endl;  void show(); void tick(); void run();time:time()  h = 0; m = 0; s = 2;time:time(int h, int m, int s)  this->h = h; this->m = m; this->s = s;void time:show()  cout << h << ":" << m &l

6、t;< ":" << s << endl; fflush(stdout);void time:tick()  sleep(1000); if (-this->s < 0)   this->s = 59;  if (-this->m < 0)    this->m = 59;   if (-this->h < 0)    this-

7、>m = 0;   void time:run()  while (this->h > 0 | this->m > 0 | this->s > 0)   show();  tick(); int main()  time t(11, 20, 5); t.run(); return 0;三、项目名称（继承与派生）：1、已知类cstring和main()函数-必做题class cstringchar *str;int size;publi

8、c:.;（1）、主程序如下：void main()cstring s1(“is a wondful ”);/调用有字符串参数的构造函数cstring s2(“programming language!”);cstring s3; /调用没有参数的构造函数s3=”c+”+(s1+s2) /调用复制（拷贝）构造函数，同时还要对“”进行重载试编写一个程序，以实现main()函数中的字符串拼接。输出结果如下：程序如下：#include<iostream>#include<string.h>#include<assert.h>using namespace std;c

9、lass cstringpublic: cstring(); cstring(const char* str); cstring(const cstring &other); cstring();public: cstring operator+(const char *lpsztext); cstring operator+(const cstring &other); cstring& operator+=(const char *lpsztext); cstring& operator+=(const cstring &other); cstrin

10、g& operator=(const char *lpsztext); cstring& operator=(const cstring &other); friend ostream & operator << (ostream &os, cstring &str) os<<str.str; return os; friend istream & operator >> (istream &is, cstring &str) char buf 1024 = 0 ; is>>

11、buf; str = buf; return is; private: char *str; int size; cstring:cstring() size = 140; str = new charsize; memset(str, 0, sizeof(char)*size);cstring:cstring(const char* str) assert(str); size = strlen(str)*2 + 1; this->str = new charsize; memcpy(this->str, str, strlen(str) + 1);cstring:cstring

12、(const cstring &other) size = other.size; this->str = new charsize; memcpy(this->str, other.str, size);cstring:cstring() if(str) delete str; size = 0;cstring cstring:operator+(const char *lpsztext) int nlen = strlen(lpsztext) + strlen(this->str) + 1; cstring str = *this; str += lpsztext

13、; return str;cstring cstring:operator+(const cstring &other) int nlen = strlen(other.str) + strlen(this->str) + 1; cstring str = *this; str += other.str; return str;cstring& cstring:operator+=(const char *lpsztext) int nlen = strlen(lpsztext) + 1; if (this->size < strlen(this->st

14、r) + nlen) this->str = (char*)realloc(this->str, (this->size) = 2 * (strlen(this->str) + nlen); strcat(this->str, lpsztext); return *this;cstring& cstring:operator+=(const cstring &other) int nlen = strlen(other.str) + 1; if (this->size < strlen(this->str) + nlen) thi

15、s->str = (char*)realloc(this->str, (this->size) = 2 * (strlen(this->str) + nlen); strcat(this->str, other.str); return *this;cstring& cstring:operator=(const char *lpsztext) int nlen = strlen(lpsztext) + 1; if(this->size < nlen) delete this->str; this->size = nlen * 2;

16、 this->str = new charsize; strcpy(this->str, lpsztext); return *this;cstring& cstring:operator=(const cstring &other) int nlen = strlen(other.str) + 1; if (this->size < nlen) delete this->str; this->size = nlen * 2; this->str = new charsize; strcpy(this->str, other.st

17、r); return *this;cstring operator+(const char *lpsztext, cstring &other) cstring str = lpsztext; str += other; return str;int main() cstring s1(" is a wondful "); cstring s2(" programming languages! "); cstring s3; s3 = "c+" + s1 + s2; cout<<s1<<endl; co

18、ut<<s2<<endl; cout<<s3<<endl; return 0;（2）、主程序如下： void main()cstring s1(“c+ is a wondful ”);/调用有字符串参数的构造函数cstring s2(“programming ”);cstring s3; /调用没有参数的构造函数s3= (s1+s2)+ “language!”/调用复制（拷贝）构造函数，同时还要对“”进行重载试编写一个程序，以实现main()函数中的字符串拼接。输出结果如（1）的结果程序如下：（前面相同）int main() cstring s1

19、(" c+ is a wondful "); cstring s2(" programming"); cstring s3; s3 = (s1 + s2)+" language! " cout<<s1<<endl; cout<<s2<<endl; cout<<s3<<endl; return 0;运行结果：（3）、主程序如下： void main()cstring s1(“c+ is a wondful ”);/调用有字符串参数的构造函数cstring s2( “

20、language!”);cstring s3; /调用没有参数的构造函数s3= s1+ “programming ”+s2 /调用复制（拷贝）构造函数，同时还要对“”进行重载试编写一个程序，以实现main()函数中的字符串拼接。输出结果如（1）的结果程序如下：int main() cstring s1(" c+ is a wondful "); cstring s2(" language!"); cstring s3; s3 = (s1 + "programming" + s2); cout<<s1<<endl

21、; cout<<s2<<endl; cout<<s3<<endl; return 0;运行结果：3、写一个程序，定义抽象基类cshape，由它派生出5个派生类：ccircle(圆形)，csquare(正方形)，crectangle(矩形)，ctrapezoid(梯形)，ctriangle(三角形)，用一个虚函数分别计算几种图形的面积，并求它们的和。要求用基类指针数组，使它每一个元素指向一个派生类对象。程序如下：#include <iostream.h>class cshapepublic: virtual double getarea

22、() = 0;class ccircle:public cshapepublic: ccircle(double r):m_r(r) virtual double getarea() return (3.14 *m_r * m_r); private: double m_r;class csquare:public cshapepublic: csquare(double a):m_a(a) virtual double getarea() return (m_a * m_a); private: double m_a;class crectangle:public cshapepublic:

23、 crectangle(double a, double b):m_a(a), m_b(b) virtual double getarea() return (m_a * m_b); private: double m_a; double m_b;class ctrapeziod:public cshapepublic: ctrapeziod(double a, double b, double h):m_a(a), m_b(b), m_h(h) virtual double getarea() return (m_a + m_b) * m_h / 2); private: double m_

24、a; double m_b; double m_h;class ctriangle:public cshapepublic: ctriangle(double a, double h):m_a(a), m_h(h) virtual double getarea() return (m_a * m_h / 2); private: double m_h; double m_a;int main() cshape* pshape5; pshape0 = new ccircle(8); pshape1 = new ctriangle(4,5); pshape2 = new csquare(5); p

25、shape3 = new ctrapeziod(55,45,7); pshape4 = new crectangle(4,55); cout<<"圆形面积:"<<pshape0->getarea()<<endl; cout<<"三角形面积:"<<pshape1->getarea()<<endl; cout<<"正方形面积:"<<pshape2->getarea()<<endl; cout<<"梯琘形面积:"<<pshape3->getarea()<<endl; cout<<"长方形面积:"<<pshape4->getarea()<<endl; double sum = 0; cout<<"面积总和为:" fo