实验3 类和对象_基础编程2
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", "Alice@hotmail.com"); 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:
