内容发布更新时间 : 2024/12/23 16:18:34星期一 下面是文章的全部内容请认真阅读。
public:
SimpleCircle(); SimpleCircle(int);
SimpleCircle(const SimpleCircle &); ~SimpleCircle() {}
void SetRadius(int); int GetRadius()const;
private: int *itsRadius; };
SimpleCircle::SimpleCircle() {
itsRadius = new int(5); }
SimpleCircle::SimpleCircle(int radius) {
itsRadius = new int(radius); }
SimpleCircle::SimpleCircle(const SimpleCircle & rhs) {
int val = rhs.GetRadius(); itsRadius = new int(val); }
int SimpleCircle::GetRadius() const {
return *itsRadius; }
int main() {
SimpleCircle CircleOne, CircleTwo(9);
cout << \cout << \return 0; }程序运行输出: CircleOne: 5 CircleTwo: 9
6-21 编写一个函数,统计一个英文句子中字母的个数,在主程序中实现输入、输出。
解: 源程序:
#include
int count(char *str) {
int i,num=0;
for (i=0; str[i]; i++) {
if ( (str[i]>='a' && str[i]<='z') || (str[i]>='A' && str[i]<='Z') ) num++; }
return num; }
void main() {
char text[100];
cout << \输入一个英语句子:\gets(text);
cout << \这个句子里有\个字母。\endl; }
程序运行输出: 输入一个英语句子: It is very interesting! 这个句子里有19个字母。
6-22 编写函数int index(char *s, char *t),返回字符串t 在字符串s中出现的最左边的位置,如果在s中没有与t匹配的子串,就返回-1。 解: 源程序:
#include
int index( char *s, char *t) {
int i,j,k;
for(i = 0; s[i] != '\\0'; i++) {
for(j = i, k = 0; t[k] != '\\0' && s[j] == t[k]; j++, k++) ;
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if (t[k] =='\\0') return i; }
return -1; }
void main() { int n;
char str1[20],str2[20]; cout << \输入一个英语单词:\cin >> str1;
cout << \输入另一个英语单词:\cin >> str2;
n = index(str1,str2); if (n > 0)
cout << str2 << \在\中左起第\<< \个位置。\else
cout << str2 << \不在\中。\}
程序运行输出:
输入一个英语单词:abcdefgh 输入另一个英语单词:de
de在abcdefghijk中左起第4个位置。
6-23 编写函数reverse(char *s)的倒序递归程序,使字符串s倒序。 解: 源程序:
#include
void reverse(char *s, char *t) { char c; if (s < t) { c = *s; *s = *t; *t = c;
reverse(++s, --t); } }
void reverse( char *s)
{
reverse(s, s + strlen(s) - 1); }
void main() {
char str1[20];
cout << \输入一个字符串:\cin >> str1;
cout << \原字符串为:\reverse(str1);
cout << \倒序反转后为:\}
程序运行输出:
输入一个字符串:abcdefghijk 原字符串为:abcdefghijk 倒序反转后为:kjihgfedcba
6-24 设学生人数N=8,提示用户输入N个人的考试成绩,然后计算出平均成绩,显示出来。 解: 源程序:
#include
#define N 8
float grades[N]; //存放成绩的数组
void main() { int i;
float total,average;
//提示输入成绩
for(i = 0; i < N; i++ ) {
cout << \cin >> grades[i]; }
total = 0;
for (i = 0; i < N; i++) total += grades[i]; average = total / N;
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cout << \}
程序运行输出: Enter grade #1: 86 Enter grade #2: 98 Enter grade #3: 67 Enter grade #4: 80 Enter grade #5: 78 Enter grade #6: 95 Enter grade #7: 78 Enter grade #8: 56
Average grade: 79.75
6-25 设计一个字符串类MyString,具有构造函数、析构函数、拷贝构造函数,重载运算符+、=、+=、[],尽可能地完善它,使之能满足各种需要。(运算符重载功能为选做,参见第8章) 解:
#include
class MyString { public: MyString();
MyString(const char *const); MyString(const MyString &); ~MyString();
char & operator[](unsigned short offset); char operator[](unsigned short offset) const; MyString operator+(const MyString&); void operator+=(const MyString&); MyString & operator= (const MyString &);
unsigned short GetLen()const { return itsLen; } const char * GetMyString() const { return itsMyString; }
private:
MyString (unsigned short); // private constructor char * itsMyString; unsigned short itsLen; };
MyString::MyString()
{
itsMyString = new char[1]; itsMyString[0] = '\\0'; itsLen=0; }
MyString::MyString(unsigned short len) {
itsMyString = new char[len+1];
for (unsigned short i = 0; i<=len; i++) itsMyString[i] = '\\0'; itsLen=len; }
MyString::MyString(const char * const cMyString) {
itsLen = strlen(cMyString); itsMyString = new char[itsLen+1];
for (unsigned short i = 0; i MyString::MyString (const MyString & rhs) { itsLen=rhs.GetLen(); itsMyString = new char[itsLen+1]; for (unsigned short i = 0; i MyString::~MyString () { delete [] itsMyString; itsLen = 0; } MyString& MyString::operator=(const MyString & rhs) { if (this == &rhs) return *this; delete [] itsMyString; itsLen=rhs.GetLen(); itsMyString = new char[itsLen+1]; for (unsigned short i = 0; i 28 return *this; } char & MyString::operator[](unsigned short offset) { if (offset > itsLen) return itsMyString[itsLen-1]; else return itsMyString[offset]; } char MyString::operator[](unsigned short offset) const { if (offset > itsLen) return itsMyString[itsLen-1]; else return itsMyString[offset]; } MyString MyString::operator+(const MyString& rhs) { unsigned short totalLen = itsLen + rhs.GetLen(); MyString temp(totalLen); for (unsigned short i = 0; i for (unsigned short j = 0; j void MyString::operator+=(const MyString& rhs) { unsigned short rhsLen = rhs.GetLen(); unsigned short totalLen = itsLen + rhsLen; MyString temp(totalLen); for (unsigned short i = 0; i for (unsigned short j = 0; j int main() { MyString s1(\ cout << \ char * temp = \s1 = temp; cout << \ char tempTwo[20]; strcpy(tempTwo,\s1 += tempTwo; cout << \cout << \ cout << \s1[4]='x'; cout << \ cout << \ MyString s2(\MyString s3; s3 = s1+s2; cout << \ MyString s4; s4 = \ cout << \return 0; } 程序运行输出: S1: initial test S1: Hello World tempTwo: ; nice to be here! S1: Hello World; nice to be here! S1[4]: o S1: Hellx World; nice to be here! S1[999]: ! S3: Hellx World; nice to be here! Another myString S4: Why does this work? 6-26 编写一个3×3矩阵转置的函数,在main()函数中输入数据。 解: #include 29 void move (int matrix[3][3]) { int i, j, k; for(i=0; i<3; i++) for (j=0; j k = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = k; } } void main() { int i, j; int data[3][3]; cout << \输入矩阵的元素\for(i=0; i<3; i++) for (j=0; j<3; j++) { cout << \第\行第\<<\个元素为:\cin >> data[i][j]; } cout << \输入的矩阵的为:\for(i=0; i<3; i++) { for (j=0; j<3; j++) cout << data[i][j] << \cout << endl; } move(data); cout << \转置后的矩阵的为:\for(i=0; i<3; i++) { for (j=0; j<3; j++) cout << data[i][j] << \cout << endl; } } 程序运行输出: 输入矩阵的元素 第1行第1个元素为:1 第1行第2个元素为:2 第1行第3个元素为:3 第2行第1个元素为:4 第2行第2个元素为:5 第2行第3个元素为:6 第3行第1个元素为:7 第3行第2个元素为:8 第3行第3个元素为:9 输入的矩阵的为: 1 2 3 4 5 6 7 8 9 转置后的矩阵的为: 1 4 7 2 5 8 3 6 9 6-27 编写一个矩阵转置的函数,矩阵的维数在程序中由用户输入。 解: #include void move (int *matrix ,int n) { int i, j, k; for(i=0; i k = *(matrix + i*n + j); *(matrix + i*n + j) = *(matrix + j*n + i); *(matrix + j*n + i) = k; } } void main() { int n, i, j; int *p; cout << \请输入矩阵的维数:\cin >> n; p = new int[n*n]; cout << \输入矩阵的元素\for(i=0; i cout << \第\行第\<<\个元素为:\cin >> p[i*n + j]; } 30