C--实验报告剖析

1、实验1-1过程化编程【实验目的】理解、掌握过程化编程程序设计思想。【实验内容】1. 程序填空，练习类、对象、继承的定义和实现方法。2. 根据程序运行结果，补充完整程序。【实验要求】我们在进行英语阅读的时候，会发现一个有趣的现象：有些字串是左右对称的，如madam我们把这种字串称为symmetry text即“对称文”。现在有若干航字串，每一行可以由数字、标点符号、空格符以及英文字符（包括大小写）组成。要你帮忙编程判断是否是对称文，否则，就不能最大限 度地发现有趣现象了。输入说明每个字串为一行，每行结束以回车符为标志，可能有上百上千行业说不定。 当字串为“ 000000时，输入结束。英文字符不区

2、分大小写，即 Madams为对称文。不要忘了 “（”与“ ） 也是互为对称的。输出说明如果是对称文，则输出“ Symmetry”，否则输出“ Not symmetry”。每个结 论占一行。Maiain*-1-madarit- kkghkkhgOOOOOM图 1 symmetry.i n: cht)6ezv_07 STnmttipSynfimetn-Not syninietn-，图 2 symmetry.out【程序代码】#inelude #inelude using namespaee std;bool isMatch( strings);int main() stri ng s;while

3、(1) cin s;if (pare( 000000) = 0) break;if (isMatch(s) cout Symmetry endl;else cout Not symmetry endl;return 0;bool isMatch( strings) int len = s.len gth();for (int i = 0; i= a & i = A &s len - i - 1 = A & i = a & len - i - 1=z ) if (si != (s len - i - 1 - ( a- A )return false ;elseif(si=&slen -

4、i -1=)continue ;elseif(si=)continue ;elseif(si=&j len -i -1=T)continue ;elseif(si=(&j len -i -1=)continue ;elseif(si !=s len-i- 1 ) return falsereturn true【运行结果】E；vjsual stdioI趕XProjectiDebugPfojecti exe 口 1Ulifinc try vn4Hi(flanling 121 yn11 Not; s yntne c rIkicyliikkhyHot 5 yrnfietrV貝财拼音输人迭全2w图3实验

5、一运行结果【实验目的】理解面向对象的的程序设计思想。【实验内容】定义一个时间类Time,能提供和设置由时、分、秒组成的时间，并编出应 用程序，要求包括定义时间对象，设置时间，输出该对象提供的时间。并请 类定义作为界面，用多文件结构实现之。【程序代码】/Time.h#include class Timepublic :int h;int m;int s;void inputT();void changeT();void outputT();/Time.cpp#include Time.h#include void Time:inputT()begin:int a, b, c;std:cout a

6、 b c;if (c 60 | c 0)std:cout 60 | b 0)std:cout 24 | a 0)std:cout Wrong time!Please set again!n ; goto begin;else if (a = 24)if (b!=0 | c!=0)std:cout Wrong time!Please set again!n goto begin;elseh = a;m = b;s = c;elseh = a;m = b;s = c;void Time:changeT()char p;std:cout p;if (p = n | p = N )std:cout T

7、hank you for your using!nelsebegin1:int a, b, c;std:cout a b c;if (c 60 | c 0)std:cout 60 | b 0)std:cout 24 | a 0)std:cout Wrong time!Please set again!n;goto begin1;else if (a = 24)if (b != 0 | c != 0)std:cout Wrong time!Please set again!n goto begin1; else h = a; m = b; s = c;else h = a; m = b; s =

8、 c;void Time:outputT()std:cout Output time(H:M:S)n h : m : s; /testmai n.cpp#in clude Time.h#include void main(void )Time time1;time1.i nputT();time1.outputT();time1.cha ngeT();time1.outputT();【运行结果】uus亡t$Projcc15Dpb gInput iIK 1H 1Output bind CH101M ：lDo if AUthAD 时 tVJnfiit t jikeIQ 1 i粘Urcng昭匕 se

9、t again*Input clrneM :N：&图4实验二运行结果实验 3 面向对象 程序设计( 2)【实验要求】改写程序 f0815.cpp ，使之含有构造函数，拷贝构造函数和析构函数。并 对主函数和矩阵向量的乘法也进行改写。对于第91和 92行，合并成“multiply(ve,ma).display();”使之不会产生内存泄漏。【实验程序】/ 实验三/ 改写 f0815.cpp#include #include #include using namespace std;class Vectorint * v; / 指向一个数组，表示向量int sz;public :int size()

10、return sz; Vector( int );Vector( const Vector & s);int & operator (int );void display();Vector();Vector :Vector( int s)sz = s;if (s = 0)cerr bad Vector size.n ;exit(1);v = new i nt s;Vector :Vector( const Vector & s)int i;sz = s.sz;v = new i nt sz;for (i = 0; isz; i+)vi = s.vi;Vector :Vector()delete

11、 v;int & Vector : operator (int i )/ 引用返回的目的是返回值可以做左值if (i = sz)cerr Vector index out of rang.n ; exit(1);return v i ;void Vector :display()int i;for (i = 0; isz; +i)cout vi ;cout n ;class Matrixint * m;int szl, szr;public :Matrix( int , int );Matrix( constMatrix & m);Matrix();int sizeL() returnszl;

12、int sizeR() returnszr;int & elem( int, int );Matrix :Matrix( int i , int j )szl = i ; szr = j ;if (i = 0 | j = 0)cerr bad Matrix size.n exit(1);m = new i nt i *j ;Matrix :Matrix( const Matrix & s)int i, j;szl = s.szl;szr = s.szr; m = new i nt szl*szr;for (i = 0; iszl; i+)for (j = 0; jszr; j+)mi*szr

13、+ j = s.mi*szr + j;Matrix :Matrix()delete m;int & Matrix :elem( int i, int j )/ 引用返回值的目的是可以做左值if (i0 | szl = i | j 0 | szr = j)cerr Matrix index out of range.n ;exit(1);return mi *szr + j ;Vector multiply( Matrix & m, Vector & v)/ 矩阵乘向量int i, j;Matrix me(m);Vector va( v);if ( m.sizeR() != v.size()ce

14、rr bad multiply Matrix with vector.n ; exit(1);Vector r( m.sizeL(); / 创建一个存放结果的空向量for (i = 0; ime.sizeL(); i+)r i = 0;for (j = 0; j x y; Matrix ma(x, y);for (i = 0; ix; +i)for (j = 0; j ma.elem(i, j);in x;Vector ve(x);for (i = 0; i ve i ;Matrix me(ma);Vector va(ve);multiply(ma, ve).display();【实验结果】s

15、iC: Wi n dotene继续图5实验三运行结果实验 4 面向对象 程序设计( 3)【实验要求】请在程序 f0904.cpp 中 的日期类的基础上，实现一个可以进 行加天数操作获 得另一个日期，以及进行日期减日期操作获得相隔天数的日期类，并进行应 用 程序设计：创建 2015.8.21和 2008.8.8两个日期，并计算中间相隔的天数，前者加上 300 天会是什么日子呢？【实验程序】一、头文件部分#pragma once/class Date with year-month-day version#include #include using namespace std; class Da

16、te int year, month, day;Date( int n = 1) i2ymd( n);int ymd2i() const ;void i2ymd( int n);static const int tians;public :Date( const string & s);Date( int y,int m,int d):year( y),month( m),day( d) Date operator+ (int n) const return Date (ymd2i() + n);Date& operator+= ( int n) i2ymd(ymd2i() + n); ret

17、urn *this ;Date& operator+ () return *this += 1;int operator- ( Date & d) const return ymd2i() - d.ymd2i();bool isLeapYear() const return !(year %4) & (year % 100) | !(year % 400);friend ostream & operator ( ostream & o, const Date& d);、函数定义/Date.cpp/Class Date with year-month-day Version#include Da

18、te.h#include #include #include using namespace std;const int Date :tians= 0,31,59,90,120,151,181,212,243,273,304,334const int Y400 = 146097; /number of days of 400 yearsconst int Y100 = 36524; /number of days of 100 yearsconst int Y4 = 1461; /number of days of 4 yearsDate:Date( const string & s) yea

19、r = atoi( s.substr(0, 4).c_str(); month = atoi( s .substr(5, 2).c_str(); day = atoi( s.substr(8, 2).c_str();void Date:i2ymd( int absDay) year = absDay / Y400 * 400;int y = absDay%Y400/;/ 被 400年除得的天数if (y = Y400 - 1) month = 12, day = 30; return ;year += y / Y100 * 100; y %= Y100;year += y / Y4 * 4;

20、y%=Y4;if (y = Y4 - 1) month = 12, day = 30; return ;year += y / 365; y %= 365;if (y = 0) month = 12, day = 31; return ;year+;bool leap = isLeapYear();for (month = 1; monthtiansmonth + (month = 2 & leap); day = y - tiansmonth - 1;int Date:ymd2i() const ;month+);- 1) / 100 + (year - 1) /int yearDay =

21、(year - 1) * 365 + (year - 1) / 4 - (yearretur n yearDay + tia nsm onth - 1 + (isLeapYear() & month 2) + day;ostream & operator* (ostream & o, const Date& d) return o setfill( O ) setw(4) d.year - setw(2) d.m on th - setw(2) d.day setfill();三、testmain/testmai n.cpp/us ing Dateclass#include Date.h#in

22、clude using namespace std;int main() Date d1(2005, 8, 21);Date d2(2008, 8, 8); d2 - d1 ncout 2005.8.21与2008.8 , 8中间相隔的天数是:cout 2005.8.21加上300天是: d1 + 300 n【程序结果】实验 5 面向对象 程序设计( 4)【实验要求】在上题 Date 类 的基础上，继承一个 WDate 类，它包含了星期几信 息，因 而，显示日期的成员要做修改，应同时显示星期几。另外，还要增加获得星 期 几的成员。想一想，类中数据成员置年、月、日好呢，还是绝对天数好呢？进而进行

23、应用程序设计：创建 2005.8.21和 2008.8.8两个日期，分别显示这两个日期。【实验程序】一、头文件 Date.h#pragma once/Date.h/class Date with year-month-day version#include #include using namespace std;class Date int year, month, day;static const int tians;protected :Date( int n = 1) i2ymd( n);int ymd2i( ) const ;void i2ymd( int n);public :Da

24、te( const string & s );Date( int y, int m, int d) :year( y), month( m), day( d) Date operator+ (int n) const return Date (ymd2i() + n);Date& operator+= (int n) i2ymd(ymd2i() + n);return *this ;Date& operator+ () return *this += 1;int operator- ( Date& d) const return ymd2i() - d.ymd2i();bool isLeapY

25、ear() const return !(year %4) & (year % 100) | !(year % 400);friend ostream & operator ( ostream & o, const Date& d);二、头文件 WDate.h#pragma once/WDate.h#include Date.h#include #include using namespace std;class WDate : public Date protected :WDate(int n = 1) : Date(n) WDate(const Date& d) : Date( d) p

26、ublic :WDate(const string & s) : Date( s) WDate(int y, int m, int d) : Date( y, m, d) WDate operator+ (int n) const return Date: operator+ (n); WDate& operator+= (int n) Date: operator+= (n); return *this ; WDate& operator+ () return *this += 1;int getWeekDay() return ymd2i() % 7; /0:Sunday,1:Monday

27、,etc. int operator- ( WDate& wd) const return ymd2i() - wd.ymd2i(); friend ostream & operator ( ostream & o, const WDate& wd);三、Date.cpp/Date.cpp/Class Date with year-month-day Version#include Date.h#include #include #include using namespace std;const int Date :tians= 0,31,59,90,120,151,181,212,243,

28、273,304,334;const int Y400 = 146097; /number of days of 400 yearsconst int Y100 = 36524; /number of days of 100 years const int Y4 = 1461; /number of days of 4 yearsDate:Date( const string & s) year = atoi( s.substr(0, 4).c_str(); month = atoi( s .substr(5, 2).c_str();day = atoi( s.substr(8, 2).c_st

29、r();void Date:i2ymd( int absDay) year = absDay / Y400 * 400;int y = absDay%Y400/;/ 被 400年除得的天数 if (y = Y400 - 1) month = 12, day = 30;returnyear += y / Y100 * 100;y %= Y100;year += y / Y4 * 4;y %= Y4;if (y = Y4 - 1) month = 12, day = 30;return ;year += y / 365;y %= 365;if (y = 0) month = 12, day = 3

30、1;return ;year+;bool leap = isLeapYear();for (month = 1; monthtiansmonth + (month = 2 & leap); month+); day = y - tiansmonth - 1;int Date:ymd2i() const int yearDay = (year - 1) * 365 + (year - 1) / 4 - (year- 1) / 100 + (year - 1) /400;return yearDay + tiansmonth - 1 + (isLeapYear() & month 2) + day;ostream & operator ( ostream & o, const Date& d) return o setfill( 0 ) setw(4) d.year - setw(2) d.month - setw(2) d.day setfill( );四、W