task1:
button.hpp:
1 #pragma once
2
3 #include <iostream>
4 #include <string>
5
6 using std::string;
7 using std::cout;
8
9 // 按钮类
10 class Button {
11 public:
12 Button(const string &text);
13 string get_label() const;
14 void click();
15
16 private:
17 string label;
18 };
19
20 Button::Button(const string &text): label{text} {
21 }
22
23 inline string Button::get_label() const {
24 return label;
25 }
26
27 void Button::click() {
28 cout << "Button '" << label << "' clicked\n";
29 }
window.hpp:
1 #pragma once
2 #include "button.hpp"
3 #include <vector>
4 #include <iostream>
5
6 using std::vector;
7 using std::cout;
8 using std::endl;
9
10 // 窗口类
11 class Window{
12 public:
13 Window(const string &win_title);
14 void display() const;
15 void close();
16 void add_button(const string &label);
17
18 private:
19 string title;
20 vector<Button> buttons;
21 };
22
23 Window::Window(const string &win_title): title{win_title} {
24 buttons.push_back(Button("close"));
25 }
26
27 inline void Window::display() const {
28 string s(40, '*');
29
30 cout << s << endl;
31 cout << "window title: " << title << endl;
32 cout << "It has " << buttons.size() << " buttons: " << endl;
33 for(const auto &i: buttons)
34 cout << i.get_label() << " button" << endl;
35 cout << s << endl;
36 }
37
38 void Window::close() {
39 cout << "close window '" << title << "'" << endl;
40 buttons.at(0).click();
41 }
42
43 void Window::add_button(const string &label) {
44 buttons.push_back(Button(label));
45 }
task1.cpp:
1 #include "window.hpp"
2 #include <iostream>
3
4 using std::cout;
5 using std::cin;
6
7 void test() {
8 Window w1("new window");
9 w1.add_button("maximize");
10 w1.display();
11 w1.close();
12 }
13
14 int main() {
15 cout << "用组合类模拟简单GUI:\n";
16 test();
17 }
task1运行测试结果截图:
问题1：定义了Button和Window两个类，使用了标准库的string和vector两个类
Button和string之间组合，Window和vector、Button组合。
问题2：不适合，添加const或inline是将类的元素固定或者内联，改变其他成员函数类型会导致程序出错。
问题3：将s数组前40个单元存放“*”。
task2：
1 #include <iostream>
2 #include <vector>
3
4 using namespace std;
5
6 void output1(const vector<int> &v) {
7 for(auto &i: v)
8 cout << i << ", ";
9 cout << "\b\b \n";
10 }
11
12 void output2(const vector<vector<int>> v) {
13 for(auto &i: v) {
14 for(auto &j: i)
15 cout << j << ", ";
16 cout << "\b\b \n";
17 }
18 }
19
20 void test1() {
21 vector<int> v1(5, 42);
22 const vector<int> v2(v1);
23
24 v1.at(0) = -999;
25 cout << "v1: "; output1(v1);
26 cout << "v2: "; output1(v2);
27 cout << "v1.at(0) = " << v1.at(0) << endl;
28 cout << "v2.at(0) = " << v2.at(0) << endl;
29 }
30
31 void test2() {
32 vector<vector<int>> v1{{1, 2, 3}, {4, 5, 6, 7}};
33 const vector<vector<int>> v2(v1);
34
35 v1.at(0).push_back(-999);
36 cout << "v1: \n"; output2(v1);
37 cout << "v2: \n"; output2(v2);
38
39 vector<int> t1 = v1.at(0);
40 cout << t1.at(t1.size()-1) << endl;
41
42 const vector<int> t2 = v2.at(0);
43 cout << t2.at(t2.size()-1) << endl;
44 }
45
46 int main() {
47 cout << "测试1:\n";
48 test1();
49
50 cout << "\n测试2:\n";
51 test2();
52 }
task2运行测试结果截图:
问题1：
vector<int> v1(5, 42); 创建了一个名为 v1 的动态数组其初始化一个可以容纳 5 个元素的向量，并 且所有初始元素值都设置为 42。
vector<int> v2(v1); 将v1动态数组内容复制到v2动态数组中。
v1.at(0) = -999; 将v1中第一个数赋值为-999。问题2：
vector<vector<int>> v1{{1, 2, 3}, {4, 5, 6, 7}}; 定义了一个二维动态数组vector<vector<int>> v1，它包含两个一维向量，每个一维向量分别存储整数。第一个向量{1, 2, 3}有三个元素，第二个向量{4, 5, 6, 7}有四个元素。
vector<vector<int>> v2(v1);将v1动态数组内容复制到v2动态数组中。
v1.at(0).push_back(-999);将v1动态数组第一个元素后插入-999。
问题3：
vector<int> t1 = v1.at(0);将v1向量中索引为0的元素赋值给t1
cout << t1.at(t1.size()-1) << endl;输出t1向量的最后一个元素
const vector<int> t2 = v2.at(0);将v2向量中索引为0的子向量赋值给t2
cout << t2.at(t2.size()-1) << endl;输出t2向量的最后一个元素
问题4：
标准库模板类vector内部封装的复制构造函数，其实现机制是深复制还是浅复制？ 深复制。 模板类vector的接口at(), 是否至少需要提供一个const成员函数作为接口？ 是。 task3: vectorInt.hpp: 1 #pragma once
2
3 #include <iostream>
4 #include <cassert>
5
6 using std::cout;
7 using std::endl;
8
9 // 动态int数组对象类
10 class vectorInt{
11 public:
12 vectorInt(int n);
13 vectorInt(int n, int value);
14 vectorInt(const vectorInt &vi);
15 ~vectorInt();
16
17 int& at(int index);
18 const int& at(int index) const;
19
20 vectorInt& assign(const vectorInt &v);
21 int get_size() const;
22
23 private:
24 int size;
25 int *ptr; // ptr指向包含size个int的数组
26 };
27
28 vectorInt::vectorInt(int n): size{n}, ptr{new int[size]} {
29 }
30
31 vectorInt::vectorInt(int n, int value): size{n}, ptr{new int[size]} {
32 for(auto i = 0; i < size; ++i)
33 ptr[i] = value;
34 }
35
36 vectorInt::vectorInt(const vectorInt &vi): size{vi.size}, ptr{new int[size]} {
37 for(auto i = 0; i < size; ++i)
38 ptr[i] = vi.ptr[i];
39 }
40
41 vectorInt::~vectorInt() {
42 delete [] ptr;
43 }
44
45 const int& vectorInt::at(int index) const {
46 assert(index >= 0 && index < size);
47
48 return ptr[index];
49 }
50
51 int& vectorInt::at(int index) {
52 assert(index >= 0 && index < size);
53
54 return ptr[index];
55 }
56
57 vectorInt& vectorInt::assign(const vectorInt &v) {
58 delete[] ptr; // 释放对象中ptr原来指向的资源
59
60 size = v.size;
61 ptr = new int[size];
62
63 for(int i = 0; i < size; ++i)
64 ptr[i] = v.ptr[i];
65
66 return *this;
67 }
68
69 int vectorInt::get_size() const {
70 return size;
71 }
task3.cpp:
1 #include "vectorInt.hpp"
2 #include <iostream>
3
4 using std::cin;
5 using std::cout;
6
7 void output(const vectorInt &vi) {
8 for(auto i = 0; i < vi.get_size(); ++i)
9 cout << vi.at(i) << ", ";
10 cout << "\b\b \n";
11 }
12
13
14 void test1() {
15 int n;
16 cout << "Enter n: ";
17 cin >> n;
18
19 vectorInt x1(n);
20 for(auto i = 0; i < n; ++i)
21 x1.at(i) = i*i;
22 cout << "x1: "; output(x1);
23
24 vectorInt x2(n, 42);
25 vectorInt x3(x2);
26 x2.at(0) = -999;
27 cout << "x2: "; output(x2);
28 cout << "x3: "; output(x3);
29 }
30
31 void test2() {
32 const vectorInt x(5, 42);
33 vectorInt y(10, 0);
34
35 cout << "y: "; output(y);
36 y.assign(x);
37 cout << "y: "; output(y);
38
39 cout << "x.at(0) = " << x.at(0) << endl;
40 cout << "y.at(0) = " << y.at(0) << endl;
41 }
42
43 int main() {
44 cout << "测试1: \n";
45 test1();
46
47 cout << "\n测试2: \n";
48 test2();
49 }
task3运行测试结果截图:
问题1：vectorInt类中，复制构造函数（line14)的实现，是深复制还是浅复制？
深复制
问题2：vectorInt类中，这两个at()接口，如果返回值类型改成int而非int&（相应地，实现部分也
同步修改），测试代码还能正确运行吗？如果把line18返回值类型前面的const掉，针对这个测试
代码，是否有潜在安全隐患？尝试分析说明。
不能正常运行，存在隐患
问题3：vectorInt类中，assign()接口，返回值类型可以改成vectorInt吗？你的结论，及，原因分
析。
可以，代码可运行，但是会占用额外内存。
task4:
matrix.hpp:
1 #pragma once
2
3 #include <iostream>
4 #include <cassert>
5
6 using std::cout;
7 using std::endl;
8
9 // 类Matrix的声明
10 class Matrix {
11 public:
12 Matrix(int n, int m); // 构造函数，构造一个n*m的矩阵, 初始值为value
13 Matrix(int n); // 构造函数，构造一个n*n的矩阵, 初始值为value
14 Matrix(const Matrix &x); // 复制构造函数， 使用已有的矩阵X构造
15 ~Matrix();
16
17 void set(const double *pvalue); // 用pvalue指向的连续内存块数据按行为矩阵赋值
18 void clear(); // 把矩阵对象的值置0
19
20 const double& at(int i, int j) const; // 返回矩阵对象索引(i,j)的元素const引用
21 double& at(int i, int j); // 返回矩阵对象索引(i,j)的元素引用
22
23 int get_lines() const; // 返回矩阵对象行数
24 int get_cols() const; // 返回矩阵对象列数
25
26 void display() const; // 按行显示矩阵对象元素值
27
28 private:
29 int lines; // 矩阵对象内元素行数
30 int cols; // 矩阵对象内元素列数
31 double *ptr;
32 };
33
34 // 类Matrix的实现：待补足
35 Matrix::Matrix(int n, int m) : lines(n), cols(m), ptr(new double[lines * cols]) {
36 clear();
37 }
38
39 Matrix::Matrix(int n) : Matrix(n, n) {}
40
41 Matrix::Matrix(const Matrix& x) : lines(x.lines), cols(x.cols), ptr(new double[lines * cols]) {
42 for (int i = 0; i < lines * cols; ++i) {
43 ptr[i] = x.ptr[i];
44 }
45 }
46
47 Matrix::~Matrix() {
48 delete[] ptr;
49 }
50
51 void Matrix::set(const double* pvalue) {
52 for (int i = 0; i < lines * cols; ++i) {
53 ptr[i] = pvalue[i];
54 }
55 }
56
57 void Matrix::clear() {
58 for (int i = 0; i < lines * cols; ++i) {
59 ptr[i] = 0;
60 }
61 }
62
63 const double& Matrix::at(int i, int j) const {
64 assert(i >= 0 && i < lines && j >= 0 && j < cols);
65 return ptr[i * cols + j];
66 }
67
68 double& Matrix::at(int i, int j) {
69 assert(i >= 0 && i < lines && j >= 0 && j < cols);
70 return ptr[i * cols + j];
71 }
72
73 int Matrix::get_lines() const {
74 return lines;
75 }
76
77 int Matrix::get_cols() const {
78 return cols;
79 }
80
81 void Matrix::display() const {
82 for (int i = 0; i < lines; ++i) {
83 for (int j = 0; j < cols; ++j) {
84 cout << at(i, j) << " ";
85 }
86 cout << endl;
87 }
88 }
task4.cpp
#include "matrix.hpp"
#include <iostream>
#include <cassert>
using std::cin;
using std::cout;
using std::endl;
const int N = 1000;
// 输出矩阵对象索引为index所在行的所有元素
void output(const Matrix &m, int index) {
assert(index >= 0 && index < m.get_lines());
for(auto j = 0; j < m.get_cols(); ++j)
cout << m.at(index, j) << ", ";
cout << "\b\b \n";
}
void test1() {
double x[1000] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int n, m;
cout << "Enter n and m: ";
cin >> n >> m;
Matrix m1(n, m); // 创建矩阵对象m1, 大小n×m
m1.set(x); // 用一维数组x的值按行为矩阵m1赋值
Matrix m2(m, n); // 创建矩阵对象m1, 大小m×n
m2.set(x); // 用一维数组x的值按行为矩阵m1赋值
Matrix m3(2); // 创建一个2×2矩阵对象
m3.set(x); // 用一维数组x的值按行为矩阵m4赋值
cout << "矩阵对象m1: \n"; m1.display(); cout << endl;
cout << "矩阵对象m2: \n"; m2.display(); cout << endl;
cout << "矩阵对象m3: \n"; m3.display(); cout << endl;
}
void test2() {
Matrix m1(2, 3);
m1.clear();
const Matrix m2(m1);
m1.at(0, 0) = -999;
cout << "m1.at(0, 0) = " << m1.at(0, 0) << endl;
cout << "m2.at(0, 0) = " << m2.at(0, 0) << endl;
cout << "矩阵对象m1第0行: "; output(m1, 0);
cout << "矩阵对象m2第0行: "; output(m2, 0);
}
int main() {
cout << "测试1: \n";
test1();
cout << "测试2: \n";
test2();
}
task4运行测试结果截图:
task5:
user.hpp:
1 #pragma once
2
3 #include <iostream>
4 #include <string>
5
6 class User {
7 private:
8 std::string name;
9 std::string password;
10 std::string email;
11
12 public:
13
14 User(const std::string& username, const std::string& userpassword = "123456", const std::string& useremail = "")
15 : name(username), password(userpassword), email(useremail) {}
16
17
18 void set_email() {
19 std::string input;
20 while (true) {
21 std::cout << "Enter email address: ";
22 std::cin >> input;
23 if (input.find('@') != std::string::npos) {
24 email = input;
25 std::cout << "email is set successfully..." << std::endl;
26 break;
27 } else {
28 std::cout << "illegal email, please re-enter email:" ;
29 std::cin>>email;
30 std::cout;
31 }
32 }
33 }
34
35
36 void change_password() {
37 std::string old_password;
38 int attempts = 1;
39 while (attempts < 3) {
40 std::cout << "Enter old password: ";
41 std::cin >> old_password;
42 if (old_password == password) {
43 std::string new_password;
44 std::cout << "Enter new password: ";
45 std::cin >> new_password;
46 password = new_password;
47 std::cout << "new password is set successfully..." << std::endl;
48 return;
49 } else {
50 std::cout << "password input error.Please re-enter again:" ;
51 attempts++;
52 }
53 }
54 std::cout << "password input error. Please try after a while." << std::endl;
55 }
56
57
58 void display() const {
59 std::cout << "name: " << name << std::endl;
60 std::cout << "password: ";
61 for (size_t i = 0; i < password.length(); ++i) {
62 std::cout << '*';
63 }
64 std::cout << std::endl;
65 std::cout << "email: " << email << std::endl;
66 }
67 };
task.cpp:
1 #include "user.hpp"
2 #include <iostream>
3 #include <vector>
4 #include <string>
5
6
7 using std::cin;
8 using std::cout;
9 using std::endl;
10 using std::vector;
11 using std::string;
12
13 void test() {
14 vector<User> user_lst;
15
16 User u1("Alice", "2024113", "[email protected]");
17 user_lst.push_back(u1);
18 cout << endl;
19
20 User u2("Bob");
21 u2.set_email();
22 u2.change_password();
23 user_lst.push_back(u2);
24 cout << endl;
25
26 User u3("Hellen");
27 u3.set_email();
28 u3.change_password();
29 user_lst.push_back(u3);
30 cout << endl;
31
32 cout << "There are " << user_lst.size() << " users. they are: " << endl;
33 for (auto& i : user_lst) {
34 i.display();
35 cout << endl;
36 }
37 }
38
39 int main() {
40 test();
41 }
task5运行测试结果截图:
task6:
date.h:
1 //date.h
2 #ifndef __DATE_H__
3 #define __DATE_H__
4 class Date { //日期类
5 private:
6 int year; //年
7 int month; //月
8 int day; //日
9 int totalDays; //该日期是从公元元年1月1日开始的第几天
10 public:
11 Date(int year, int month, int day); //用年、月、日构造日期
12 int getYear() const { return year; }
13 int getMonth() const { return month; }
14 int getDay() const { return day; }
15 int getMaxDay() const; //获得当月有多少天
16 bool isLeapYear() const { //判断当年是否为闰年
17 return year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
18 }
19 void show() const; //输出当前日期
20 //计算两个日期之间差多少天
21 int distance(const Date & date) const {
22 return totalDays - date.totalDays;
23 }
24 };
25 #endif //__DATE_H__
date.cpp:
1 #include "date.h"
2 #include <iostream>
3 #include <cstdlib>
4 using namespace std;
5 namespace { //namespace使下面的定义只在当前文件中有效
6 //存储平年中的某个月1日之前有多少天，为便于getMaxDay函数的实现，该数组多出一项
7 const int DAYS_BEFORE_MONTH[] = { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 };
8 }
9 Date::Date(int year, int month, int day) : year(year), month(month), day(day) {
10 if (day <= 0 || day > getMaxDay()) {
11 cout << "Invalid date: ";
12 show();
13 cout << endl;
14 exit(1);
15 }
16 int years = year - 1;
17 totalDays = years * 365 + years / 4 - years / 100 + years / 400
18 + DAYS_BEFORE_MONTH[month - 1] + day;
19 if (isLeapYear() && month > 2) totalDays++;
20 }
21 int Date::getMaxDay() const {
22 if (isLeapYear() && month == 2)
23 return 29;
24 else
25 return DAYS_BEFORE_MONTH[month] - DAYS_BEFORE_MONTH[month - 1];
26 }
27 void Date::show() const {
28 cout << getYear() << "-" << getMonth() << "-" << getDay();
29 }
account.h:
1 //account.h
2 #ifndef __ACCOUNT_H__
3 #define __ACCOUNT_H__
4 #include "date.h"
5 #include <string>
6 class SavingsAccount { //储蓄账户类
7 private:
8 std::string id; //帐号
9 double balance; //余额
10 double rate; //存款的年利率
11 Date lastDate; //上次变更余额的时期
12 double accumulation; //余额按日累加之和
13 static double total; //所有账户的总金额
14 //记录一笔帐，date为日期，amount为金额，desc为说明
15 void record(const Date & date, double amount, const std::string & desc);
16 //报告错误信息
17 void error(const std::string & msg) const;
18 //获得到指定日期为止的存款金额按日累积值
19 double accumulate(const Date & date) const {
20 return accumulation + balance * date.distance(lastDate);
21 }
22 public:
23 //构造函数
24 SavingsAccount(const Date & date, const std::string & id, double rate);
25 const std::string & getId() const { return id; }
26 double getBalance() const { return balance; }
27 double getRate() const { return rate; }
28 static double getTotal() { return total; }
29 //存入现金
30 void deposit(const Date & date, double amount, const std::string & desc);
31 //取出现金
32 void withdraw(const Date & date, double amount, const std::string & desc);
33 //结算利息，每年1月1日调用一次该函数
34 void settle(const Date & date);
35 //显示账户信息
36 void show() const;
37 };
38 #endif //__ACCOUNT_H__
account.cpp:
1 //account.cpp
2 #include "account.h"
3 #include <cmath>
4 #include <iostream>
5 using namespace std;
6 double SavingsAccount::total = 0;
7 //SavingsAccount类相关成员函数的实现
8 SavingsAccount::SavingsAccount(const Date & date, const string & id, double rate)
9 : id(id), balance(0), rate(rate), lastDate(date), accumulation(0) {
10 date.show();
11 cout << "\t#" << id << " created" << endl;
12 }
13 void SavingsAccount::record(const Date & date, double amount, const string & desc) {
14 accumulation = accumulate(date);
15 lastDate = date;
16 amount = floor(amount * 100 + 0.5) / 100; //保留小数点后两位
17 balance += amount;
18 total += amount;
19 date.show();
20 cout << "\t#" << id << "\t" << amount << "\t" << balance << "\t" << desc << endl;
21 }
22 void SavingsAccount::error(const string & msg) const {
23 cout << "Error(#" << id << "): " << msg << endl;
24 }
25 void SavingsAccount::deposit(const Date & date, double amount, const string & desc) {
26 record(date, amount, desc);
27 }
28 void SavingsAccount::withdraw(const Date & date, double amount, const string & desc) {
29 if (amount > getBalance())
30 error("not enough money");
31 else
32 record(date, -amount, desc);
33 }
34 void SavingsAccount::settle(const Date & date) {
35 double interest = accumulate(date) * rate //计算年息
36 / date.distance(Date(date.getYear() - 1, 1, 1));
37 if (interest != 0)
38 record(date, interest, "interest");
39 accumulation = 0;
40 }
41 void SavingsAccount::show() const {
42 cout << id << "\tBalance: " << balance;
43 }
task6.cpp
1 //6_25.cpp
2 #include "account.h"
3 #include <iostream>
4 using namespace std;
5 int main() {
6 Date date(2008, 11, 1); //起始日期
7 //建立几个账户
8 SavingsAccount accounts[] = {
9 SavingsAccount(date, "03755217", 0.015),
10 SavingsAccount(date, "02342342", 0.015)
11 };
12 const int n = sizeof(accounts) / sizeof(SavingsAccount); //账户总数
13 //11月份的几笔账目
14 accounts[0].deposit(Date(2008, 11, 5), 5000, "salary");
15 accounts[1].deposit(Date(2008, 11, 25), 10000, "sell stock 0323");
16 //12月份的几笔账目
17 accounts[0].deposit(Date(2008, 12, 5), 5500, "salary");
18 accounts[1].withdraw(Date(2008, 12, 20), 4000, "buy a laptop");
19 //结算所有账户并输出各个账户信息
20 cout << endl;
21 for (int i = 0; i < n; i++) {
22 accounts[i].settle(Date(2009, 1, 1));
23 accounts[i].show();
24 cout << endl;
25 }
26 cout << "Total: " << SavingsAccount::getTotal() << endl;
27 return 0;
28 }
task6运行测试结果截图:
标签：std,const,cout,对象,编程,int,实验,include,string From： https://www.cnblogs.com/pic-riy/p/18525767