1.三维坐标点（运算符重载）
题目：
设计一个三维点坐标的类Point3D，要求如下：
（1）包含三个int类型数据成员my_x, my_y, my_z；
（2）默认构造函数，默认三维点的坐标值为0, 0, 0；
（3）SetPoint函数，设置坐标值；
（4）用成员函数重载”+”运算符，实现两个Point3D类对象的加法运算；
（5）重载“+”运算符，实现整型常量和PointedD类对象的加法运算。规则是：常量值与x坐标相加；
（6）在main( ) 函数中测试类Point3D“+”和"3+pt1”的功能，并输出结果。
（7）重载“<<”运算符，实现点类对象的输出，坐标值xyz之间以逗号分隔。
ostream & operator<<(ostream &out, const Point3D &pt)
{
out<<pt.my_x<<","<<pt.my_y<<","<<pt.my_z<<endl;
return out;
}
主函数已经给出，请勿修改。
int main()
{
Point3D pt1, pt2, pt3, pt4;
pt1.SetPoint(10, 20, 3);
pt2.SetPoint(40, 50, 60);
pt3 = pt1+pt2;
cout<<“pt1:”<<pt1;
cout<<“pt2:”<<pt2;
cout<<“pt1+pt2:”<<pt3;
pt4 = 3+pt1;
cout<<“3+pt1:”<<pt4;
return 0;
}
【输入形式】
两个三维点的坐标值
【输出形式】
第一行为点pt1的坐标值；第二行为点pt2的坐标值；第三行为pt1+pt2的结果；第四行为m+pt1的结果。各坐标值之间以“,”分隔。
【样例输入】
无
【样例输出】
pt1:10,20,3
pt2:40,50,60
pt1+pt2:50,70,63
3+pt1:13,20,3
代码：
#include <iostream>
using namespace std;
class Point3D{
public:
Point3D(int x = 0,int y = 0,int z = 0):my_x(x), my_y(y), my_z(z)
{
}
void SetPoint(int x, int y, int z){
my_x = x;
my_y = y;
my_z = z;
}
Point3D operator+(const Point3D &p)const{
return Point3D(my_x + p.my_x, my_y + p.my_y, my_z + p.my_z);
}
friend Point3D operator+(Point3D p, int x);
friend Point3D operator+(int x, Point3D p);
friend ostream & operator<<(ostream &out, const Point3D &pt);
private:
int my_x, my_y, my_z;
};
ostream & operator<<(ostream &out, const Point3D &pt){
out<<pt.my_x<<","<<pt.my_y<<","<<pt.my_z<<endl;
return out;
}
Point3D operator+(Point3D p, int x){
p.my_x += x;
return p;
}
Point3D operator+(int x, Point3D p){
p.my_x += x;
return p;
}
int main(){
Point3D pt1, pt2, pt3, pt4;
pt1.SetPoint(10, 20, 3);
pt2.SetPoint(40, 50, 60);
pt3 = pt1 + pt2;
cout<<"pt1:"<<pt1;
cout<<"pt2:"<<pt2;
cout<<"pt1+pt2:"<<pt3;
pt4 = 3 + pt1;
cout<<"3+pt1:"<<pt4;
return 0;
}
2.学生类的输入输出（运算符的重载）
题目：
定义一个学生类，包括私有数据成员学号、姓名、性别、成绩和其他成员函数。请编程实现以下主函数的功能。
int main()
{
student s;
cin>>s;
cout<<s;
return 0;
}
【样例输入】
201901001 Zhang f 80
【样例输出】
201901001 Zhang f 80
代码：
#include <iostream>
#include <cstring>
using namespace std;
class Student{
char id[20];
char name[10];
char sex;
int score;
public:
Student(){
strcpy(id, "000");
strcpy(name, "000");
sex = 'm';
score = 0;
}
friend istream &operator>>(istream&in, Student &s);
friend ostream &operator<<(ostream&out, const Student &s);
};
istream &operator>>(istream&in, Student &s){
in>>s.id>>s.name>>s.sex>>s.score;
return in;
}
ostream &operator<<(ostream&out, const Student &s){
out<<s.id<<" "<<s.name<<" "<<s.sex<<" "<<s.score;
return out;
}
int main()
{
Student s;
cin>>s;
cout<<s;
return 0;
}
3.二进制类（运算符的重载）
题目：
将一个16位二进制数表示成0和1的字符序列，即用一个字符数组来存放这个二进制数。在这个类中设置两个构造函数，一个是传递整数参数的，另一个是传递字符串参数的。因为用户在创建对象时传递的二进制数，可能是以整数形式给出，也可能是以数字串形式给出，系统应该都能接受。另外有一个类型转换函数int()，用来将类类型向整型转换，即将二进制形式的类对象转换为整形数。两个重载运算符“+”，“-”，用来完成两个二进制数之间的加减运算。
class binary { //定义二进制类
char bits[16]; //二进制字模数组
public:
binary(char *); //字符串参数构造函数
binary(int); //整型参数构造函数
friend binary operator +(binary,binary); //重载“+”，友元函数
friend binary operator -(binary,binary); //重载“-”，友元函数
operator int(); //类类型转换函数，成员函数
friend ostream & operator <<(ostream &out, binary &b);//重载“<<”，以二进制形式输出
void print();//以整型形式输出
};
主函数设计如下，请勿修改：
int main(){
binary n1=“1011”;
binary n2=int(n1)+15;
binary n3=n1-binary(7);
cout<<n1<<endl;
cout<<n2<<endl;
cout<<n3<<endl;
cout<<int(n2)+5<<endl;
n2=n2-binary(5);
n2.print();
n3=n3+binary(5);
n3.print();
cout<<int(n3)-5<<endl;
return 0;
}
【样例输出】
0000000000001011
0000000000011010
0000000000000100
31
21
9
4
代码：
#include <iostream>
#include <cstring>
using namespace std;
int fun(int n){
int sum = 1;
for(int i = 0; i < n; i ++){
sum *= 2;
}
return sum;
}
class binary{
char bits[16]; //二进制字模数组
public:
binary(char *c); //字符串参数构造函数
binary(int n); //整型参数构造函数
friend binary operator + (binary &b1, binary &b2); //重载“+”，友元函数
friend binary operator - (binary &b1, binary &b2); //重载“-”，友元函数
operator int(); //类类型转换函数，成员函数
friend ostream & operator <<(ostream &out, binary &b);//重载“<<”，以二进制形式输出
void print();//以整型形式输出
};
binary::operator int(){
int num = 0;
for(int i = 15; i >= 0; i --){
num += (bits[i] - '0') * fun(15 - i); //两个字符相减，得到的就是ascll差值；
}
return num;
}
binary::binary( char *c){
for(int i = 0; i < 16; i ++)bits[i] = '0';
int m = strlen(c);
int i = 15;
for(int j = m - 1; j >= 0; j --){
bits[i] = c[j];
i --;
}
}
binary::binary(int n){
for(int i = 15; i >= 0; i --){
bits[i] = n % 2 + '0';
n /= 2;
}
}
binary operator +(binary &b1, binary &b2){
int m = int(b1) + int(b2);
return binary(m);
}
binary operator -(binary &b1, binary &b2){
int m = int(b1) - int(b2);
return binary(m);
}
ostream & operator <<(ostream &out, binary &b){
for(int i = 0; i < 16; i ++){
out<<b.bits[i];
}
out<<endl;
return out;
}
void binary::print(){
cout<<int(*this)<<endl;
}
int main(){
binary n1 = "1011";
binary n2 = int(n1)+15;
binary n3 = n1 - binary(7);
cout<<n1<<endl;
cout<<n2<<endl;
cout<<n3<<endl;
cout<<int(n2) + 5<<endl;
n2 = n2 - binary(5);
n2.print();
n3 = n3 + binary(5);
n3.print();
cout<<int(n3) - 5<<endl;
return 0;
}
4.二进制类的完善
题目：
在上一题的基础上，将 + 、 - 运算符定义为binary类的成员函数。并且重载运算符 ~ 、 & 、 | ，分别表示将二进制数按位取反、按位与及按位或。主函数设计如下，请勿修改：
int main(){
binary n1=“1011”;
binary n2=int(n1)+15;
binary n3=n1-binary(7);
cout<<n1<<endl;
cout<<n2<<endl;
cout<<n3<<endl;
binary n4=n1&n2;
binary n5=n1|n2;
binary n6=~n1;
cout<<n4<<endl;
cout<<n5<<endl;
cout<<n6<<endl;
return 0;
}
【样例输出】
0000000000001011
0000000000011010
0000000000000100
0000000000001010
0000000000011011
1111111111110100
代码：
#include <iostream>
#include <cstring>
using namespace std;
int fun(int n){
int sum = 1;
for(int i = 0; i < n; i ++){
sum *= 2;
}
return sum;
}
class binary {
char bits[16];
public:
binary(){
for(int i = 0; i < 16; i ++)bits[i] = '0';}
binary(char *c);
binary(int n);
binary operator +(binary &b1);
binary operator -(binary &b1);
binary operator &(binary &b1);
binary operator |(binary &b1);
binary operator ~();
operator int();
friend ostream & operator <<(ostream &out, binary &b);
};
binary::operator int(){
int num = 0;
for(int i = 15; i >= 0; i --){
num += (bits[i] - '0') * fun(15 - i);
}
return num;
}
binary::binary( char *c){
for(int i = 0; i < 16; i ++)bits[i] = '0';
int m = strlen(c);
int i = 15;
for(int j = m - 1; j >= 0; j --){
bits[i] = c[j];
i --;
}
}
binary::binary(int n){
for(int i = 15; i >= 0; i --){
bits[i] = n % 2 + '0';
n /= 2;
}
}
binary binary::operator +(binary &b1){
int m = int(b1) + int(*this);
return binary(m);
}
binary binary::operator -(binary &b1){
int m = int(*this) - int(b1);
return binary(m);
}
binary binary::operator &(binary &b1){
binary a;
for(int i = 0, j = 0; i < 16, j < 16; i ++, j ++){
if(b1.bits[i] == '1' && bits[j] == '1')a.bits[i] = '1';
else a.bits[i] = '0';
}
return a;
}
binary binary::operator |(binary &b1){
binary a;
for(int i = 0, j = 0; i < 16, j < 16; i ++, j ++){
if(b1.bits[i] == '0' && bits[j] == '0')a.bits[i] = '0';
else a.bits[i] = '1';
}
return a;
}
binary binary::operator ~(){
binary a;
for(int i = 0; i < 16; i ++){
if(bits[i] == '0')a.bits[i] = '1';
if(bits[i] == '1')a.bits[i] = '0';
}
return a;
}
ostream & operator <<(ostream &out, binary &b){
for(int i = 0; i < 16; i ++){
out<<b.bits[i];
}
out<<endl;
return out;
}
int main(){
binary n1 = "1011";
binary n2 = int(n1) + 15;
binary n3 = n1 - binary(7);
cout<<n1<<endl;
cout<<n2<<endl;
cout<<n3<<endl;
binary n4 = n1 & n2;
binary n5 = n1 | n2;
binary n6 = ~n1;
cout<<n4<<endl;
cout<<n5<<endl;
cout<<n6<<endl;
return 0;
}
5.图形类（继承与派生，纯虚函数，动态联编的动态性）
题目：
（1）首先设计一个抽象类Shape，包含三个公有的纯虚函数，分别实现显示数据成员的值、计算面积、计算体积的功能，函数名为ShowData()、GoArea()、GoVolume()
（2）设计两个Shape的公有派生类平面图形类、立体图形类，包含公有的数据成员面积area和体积volume，并实现基类Shape的三个纯虚函数的函数体。
（3）分别设计圆形类、椭圆类、矩形类、等腰三角形类，作为平面图形类的公有派生类。并在各个类里依次添加数据成员圆形半径、椭圆的横轴和纵轴长度、矩形的长和宽、等腰三角形的底边长和高。并实现数据的显示输出函数、实现计算面积的函数。
（4）分别设计球类、圆柱体类、长方体类，作为立体图形类的公有派生类。并在各个类里依次添加数据成员球的半径，圆柱体的底面的半径和高，长方体的长宽高。并实现数据的显示输出函数、实现计算体积的函数。
（5）计算面积的函数和计算体积的函数均需要先cin相应的数据，再才能计算，计算结果为double型，且保留2位小数，将计算结果给相应的数据成员进行赋值。函数无返回值。
（6）测试数据时，定义各种类型的对象，然后定义Shape类型的指针p，用户依次输入1、2、3…6、7，分别代表圆形、椭圆、矩形、三角形、球、圆柱体、长方体。通过使用switch语句，将对象地址赋值给p指针。然后调用该对象的计算面积或计算体积的函数，输出计算结果。为方便测试，switch仅执行一次，程序则结束。
举例：
Shape *p;
Circle c;
p=&c;
p->GoArea();
p->ShowData();
【样例输入】
5
2
【样例输出】
33.51
【样例说明】5代表球，2代表球的半径，根据体积公式4/3pirrr得到33.51
代码：
#include <iostream>
#include <iomanip>
#define PI 3.1415926 //位数要写多一点，然后控制输出位数
using namespace std;
class Shape{
public:
virtual void ShowData() = 0;
virtual void GoArea() = 0;
virtual void GoVolume() = 0;
};
class Planefigure:public Shape{
//平面图形
public:
double area;
void ShowData() {
} //不定义的话也要写一个空的；
void GoArea() {
}
void GoVolume() {
}
};
class Stereogram:public Shape{
//立体图形
public:
double v;
void ShowData() {
} //不定义也要写一个空的，不然会报错；
void GoArea() {
}
void GoVolume() {
}
};
class Circular:public Planefigure{
//圆
public:
Circular(){
int n;
cin>>n;
r = n;
}
void GoArea(){
area = PI * r * r; }
void ShowData(){
cout<< fixed << setprecision(2) << area << endl; } //setprecision（n）控制总输出n位数（包括整数部分）；用到<iomanip>;
//加上fixed是使小数位数为n；不包括整数部分；
private:
int r;
};
class Ellipse:public Planefigure{
//椭圆
public:
Ellipse(){
int m, n;
cin>>m>>n;
a = m;
b = n;
}
void GoArea(){
area = PI * a * b; }
void ShowData(){
cout<< fixed << setprecision(2) << area <<endl; }
private:
int a, b;
};
class Pyramid:public Planefigure{
//矩形
public:
Pyramid(){
int m, n;
cin>>m>>n;
a = m;
b = n;
}
void GoArea(){
area = a * b; }
void ShowData(){
cout<< fixed << setprecision(2)<<area<<endl; }
private:
int a, b;
};
class Triangle:public Planefigure{
//等腰三角形
public:
Triangle(){
int m, n;
cin>>m>>n;
botten = m;
high = n;
}
void GoArea(){
area = 1.0 / 2 * botten * high; }
void ShowData(){
cout<< fixed << setprecision(2)<<area<<endl; }
private:
int botten, high;
};
class Ball:public Stereogram{
//球
public:
Ball(){
int n;
cin>>n;
R = n;
}
void GoVolume() {
v = 4.0 / 3 * PI * R * R * R; }
void ShowData() {
cout<< fixed << setprecision(2)<<v<<endl; }
private:
int R;
};
class Cylinder :public Stereogram{
//圆柱体
public:
Cylinder(){
int m, n;
cin>>m>>n;
dr = m;
h = n;
}
void GoVolume() {
v = PI * dr * dr * h; }
void ShowData() {
cout<< fixed << setprecision(2)<<v<<endl; }
private:
int dr, h;
};
class Rectangle:public Stereogram{
// 长方体
public:
Rectangle(){
int i, j, k;
cin>>i>>j>>k;
aa = i;
bb = j;
cc = k;
}
void GoVolume() {
v = aa * bb * cc; }
void ShowData() {
cout<< fixed << setprecision(2)<<v<<endl; }
private:
int aa, bb, cc;
};
int main(){
Shape *p; int n;
cin>>n;
switch(n){
case 1: p = new Circular();p->GoArea();p->ShowData();break; //当场创建对象！！
case 2: p = new Ellipse();p->GoArea();p->ShowData();break;
case 3: p = new Pyramid();p->GoArea();p->ShowData();break;
case 4: p = new Triangle();p->GoArea();p->ShowData();break;
case 5: p = new Ball();p->GoVolume();p->ShowData();break;
case 6: p = new Cylinder();p->GoVolume();p->ShowData();break;
case 7: p = new Rectangle();p->GoVolume();p->ShowData();break;
}
return 0;
}
6.时间类的改进（运算符重载）
题目：
对前面实验写过的Time类进行修改，删去Add和Sub成员函数，通过重载“+”、“-”运算符直接进行时间的加减运算。
提示：
(1)可以用友元函数来实现“+”“-”运算符的重载。
(2)加法运算符可以是两个Time对象进行相加，也可以是一个表示秒数的int型数据加上一个Time对象，还可以是Time对象加上int型数据，得到的结果都是Time类型的对象。
(3)减法运算符可以是两个Time对象进行相减，也可以是Time对象减去一个表示秒数的int型数据，得到的结果都是Time类型的对象。
主函数设计如下，请勿修改：
int main(){
Time t1(2,34),t2,t3;
t2.SetTime(13,23,34);
cout<<“t1+t2:”;
t3=t1+t2;//两个Time类对象相加
t3.print_24();
cout<<"\nt1+65:";
t3=t1+65;//Time类对象加上65秒
t3.print_24();
cout<<"\n65+t1:";
t3=65+t1;//65秒加上Time类对象
t3.print_24();
cout<<"\nt2-t1:";
t3=t2-t1;//两个Time类对象相减
t3.print_24();
cout<<"\nt1-70:";
t3=t1-70;//Time类对象减去70秒
t3.print_24();
return 0;
}
【样例输出】
t1+t2:15:57:34
t1+65:02:35:05
65+t1:02:35:05
t2-t1:10:49:34
t1-70:02:32:50
代码：
#include <iostream>
#include <iomanip>
using namespace std;
class Time
{
int hour, minute, second;
public:
Time(int h, int m, int s = 0);
Time(int s = 0);
int SecCalc(){
return(hour * 60 + minute) * 60 + second;}
void SetTime(int h = 0, int m = 0, int s = 0);
void print_12();
void print_24();
Time operator+(Time &t)const;
Time operator-(Time &t)const;
friend Time operator+(Time &t, int n);
friend Time operator+(int n, Time &t);
friend Time operator-(Time &t, int n);
};
Time::Time(int h, int m, int s){
hour = h;
minute = m;
second = s;
}
Time::Time(int s){
hour = s / 3600;
minute = (s - (hour * 3600)) / 60;
second = s - (hour * 3600 + minute * 60);
}
void Time::SetTime(int h, int m, int s){
hour = h;
minute = m;
second = s;
}
void Time::print_12(){
if(hour <= 12)
{
cout<<setw(2)<<setfill('0')<<hour<<":"<<minute<<":"<<second<<" AM";}
if(hour > 12)
{
cout<<setw(2)<<setfill('0')<<hour-12<<":"<<minute<<":"<<second<<" PM";}
}
void Time::print_24(){
cout<<setw(2)<<setfill('0')<<hour<<":"<<setw(2)<<setfill('0')<<minute<<":"<<setw(2)<<setfill('0')<<second;
}
Time Time::operator+(Time &t)const{
Time c;
c.second = second + t.second;
if(c.second >= 60){
c.second -= 60;
c.minute += 1;
}
c.minute = minute + t.minute;
if(c.minute >= 60){
c.minute -= 60;
c.hour += 1;
}
c.hour = hour + t.hour;
if(c.hour >= 24){
c.hour -= 24;
}
return c;
}
Time Time::operator-(Time &t)const{
Time c;
if((t.hour * 60 + t.minute) * 60 + t.second >= (hour * 60 + minute) * 60 + second){
int a = ((t.hour * 60 + t.minute) * 60 + t.second) - ((hour * 60 + minute) * 60 + second);
c.hour = a / 3600; //大哥长长记性！！复杂运算记得多加括号啊！！！
c.minute = (a - (c.hour * 3600)) / 60;
c.second = a - (c.hour * 3600 + c.minute * 60);
}
else{
int d = ((hour * 60 + minute) * 60 + second) - ((t.hour * 60 + t.minute) * 60 + t.second );
c.hour = d / 3600;
c.minute = (d - (c.hour * 3600)) / 60;
c.second = d - (c.hour * 3600 + c.minute * 60);
}
return c;
}
Time operator+(Time &t, int n){
Time c;
c.second = n + t.second;
if(c.second >= 60){
c.second -= 60;
c.minute += 1;
}
c.minute += t.minute;
if(c.minute >= 60){
c.minute -= 60;
c.hour += 1;
}
c.hour += t.hour;
if(c.hour >= 24){
c.hour -= 24;
}
return c;
}
Time operator+(int n, Time &t){
Time c;
c.second = n + t.second;
if(c.second >= 60){
c.second -= 60;
c.minute += 1;
}
c.minute += t.minute;
if(c.minute >= 60){
c.minute -= 60;
c.hour += 1;
}
c.hour += t.hour;
if(c.hour >= 24){
c.hour -= 24;
}
return c;
}
Time operator-(Time &t, int n){
Time c;
if(t.second >= n){
c.second = t.second - n;
}
else{
c.second = t.second + (120 - n);
c.minute += (t.minute - 2);
if(c.minute < 0){
c.minute += 60;
c.hour -= 1;
}
c.hour += t.hour;
if(c.hour < 0){
c.hour += 24;
}
}
return c;
}
int main(){
Time t1(2, 34), t2, t3;
t2.SetTime(13, 23, 34);
cout<<"t1+t2:";
t3 = t1 + t2;//两个Time类对象相加
t3.print_24();
cout<<"\nt1+65:";
t3 = t1 + 65;//Time类对象加上65秒
t3.print_24();
cout<<"\n65+t1:";
t3 = 65 + t1;//65秒加上Time类对象
t3.print_24();
cout<<"\nt2-t1:";
t3 = t2 - t1;//两个Time类对象相减
t3.print_24();
cout<<"\nt1-70:";
t3 = t1 - 70;//Time类对象减去70秒
t3.print_24();
return 0;
}
7.运算符的重载
题目：
不允许修改主函数，请将以下代码补充完整。
class String {
char *ptr;
public:
String(char *s)
{
ptr=new char[strlen(s)+1];
strcpy(ptr, s);
}
~String() { delete []ptr; }
void print() { cout<<ptr<<endl; }
};
int main( )
{ String p1(“book”), p2(“pen”),p3(“good”),p4;
p4 = p4 = p1 ;
p3 = p1 = p2;
cout<<“p2:”;
p2.print();
cout<<“p1:”<<p1;
cout<<“p3:”<<p3;
p4+=p3;
cout<<“p4:”<<p4;
return 0;
}
【样例输入】
无
【样例输出】
p2:pen
p1:pen
p3:pen
p4:bookpen
代码：
#include <iostream>
#include <cstring>
using namespace std;
class String{
char *ptr;
public:
String(){
ptr = NULL;
}
String(const char *s)
{
ptr = new char[strlen(s) + 1];
strcpy(ptr, s);
}
~String(){
delete []ptr; }
void operator +=(const String &s);
String& operator =(const String &s);
friend ostream & operator <<(ostream &out, String &s);
void print() {
cout<<ptr<<endl; }
};
void String::operator +=(const String &s){
char a[100];
strcpy(a, ptr);
delete []ptr;
ptr = new char[strlen(a) + strlen(s.ptr) + 1];
strcpy(ptr, strcat(a, s.ptr));
}
String& String::operator =(const String &s){
if(this == &s) return *this;
delete []ptr;
ptr = new char[strlen(s.ptr) + 1];
strcpy(ptr, s.ptr);
return *this;
}
ostream & operator <<(ostream &out, String &s){
out<<s.ptr<<endl;
return out;
}
int main( ){
String p1("book"), p2("pen"), p3("good"), p4;
p4 = p4 = p1;
p3 = p1 = p2;
cout<<"p2:";
p2.print();
cout<<"p1:"<<p1;
cout<<"p3:"<<p3;
p4 += p3;
cout<<"p4:"<<p4;
return 0;
}
8.字符串排序
题目：
将若干字符串按字母顺序（由小到大）输出。主函数如下，请补充完成该程序。
int main( )
{ void sort(char *alpha[], int n );
void print(char *alpha[], int n );
char *alpha[]={“Follow me”, “Basic”, “Great Wall”, “FORTRAN”, “Computer design”};
int n=5;
//分别调用sort函数和print函数完成字符串的排序及输出。请自行练习选择排序和插入排序方法。
return 0;
}
【样例输入】
无
【样例输出】
Basic
Computer design
FORTRAN
Follow me
Great Wall
代码：
#include <iostream>
#include <cstring>
using namespace std;
int main( ){
void sort(char *alpha[], int n );
void print(char *alpha[], int n );
char *alpha[] = {
"Follow me", "Basic", "Great Wall", "FORTRAN", "Computer design"};
int n = 5;
//分别调用sort函数和print函数完成字符串的排序及输出。请自行练习选择排序和插入排序方法。
sort(alpha, 5);
print(alpha, 5);
return 0;
}
void sort(char *alpha[], int n ){
//char *alpha[]是一个装着很多“指针”的数组
for(int i = 0; i < n; i ++){
for(int j = i + 1; j < n; j ++){
if(strcmp(alpha[i], alpha[j]) > 0){
char *tem = alpha[i]; //所以直接交换指针，不能用strcpy！！
alpha[i] = alpha[j];
alpha[j] = tem;
}
}
}
}
void print(char *alpha[], int n ){
for(int i = 0; i < n; i ++){
cout<<alpha[i]<<endl;
}
}
欢迎提问，学弟学妹们加油~