实验四
task2
GradeCalc.hpp
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>
#include <iomanip>
using std::vector;
using std::string;
using std::cin;
using std::cout;
using std::endl;
class GradeCalc : public vector<int> {
public:
GradeCalc(const string &cname, int size);
void input(); //录入成绩
void output() const; //输出成绩
void sort(bool ascending = false); //排序(默认降序)
int min() const; //返回最低分
int max() const; //返回最高分
float average() const; //返回平均分
void info(); //输出课程成绩信息
private:
void compute(); //成绩统计
private:
string course_name; //课程名
int n; //课程人数
vector<int> counts = vector<int>(5, 0); //保存各分数段人数([0, 60),[60,70),[70,80),[80,90),[90,100])
vector<double> rates = vector<double>(5, 0);//保存各分数段比例
};
GradeCalc::GradeCalc(const string &cname, int size) :
course_name(cname), n{size} {}
void GradeCalc::input() {
int grade;
for (int i = 0; i < n; ++i) {
cin >> grade;
this->push_back(grade);
}
}
void GradeCalc::output() const {
for (auto ptr = this->begin(); ptr!= this->end(); ++ptr)
cout << *ptr << " ";
cout << endl;
}
void GradeCalc::sort(bool ascending) {
if (ascending)
std::sort(this->begin(), this->end());
else
std::sort(this->begin(), this->end(), std::greater<int>());
}
int GradeCalc::min() const {
return *std::min_element(this->begin(), this->end());
}
int GradeCalc::max() const {
return *std::max_element(this->begin(), this->end());
}
float GradeCalc::average() const {
return std::accumulate(this->begin(), this->end(), 0) * 1.0 / n;
}
void GradeCalc::compute() {
for (int grade : *this) {
if (grade < 60)
counts.at(0)++;
else if (grade >= 60 && grade < 70)
counts.at(1)++;
else if (grade >= 70 && grade < 80)
counts.at(2)++;
else if (grade >= 80 && grade < 90)
counts.at(3)++;
else if (grade >= 90)
counts.at(4)++;
}
for (int i = 0; i < rates.size(); ++i)
rates.at(i) = counts.at(i) * 1.0 / n;
}
void GradeCalc::info() {
cout << "课程名称:\t" << course_name << endl;
cout << "排序后成绩:\t";
sort();
output();
cout << "最高分:\t" << max() << endl;
cout << "最低分:\t" << min() << endl;
cout << "平均分:\t" << std::fixed << std::setprecision(2) <<
average() << endl;
compute();//统计各分数段人数、比例
vector<string> tmp{"[0,60)", "[60,70)", "[70,80)", "[80,90)",
"[90,100]"};
for (int i = tmp.size() - 1; i >= 0; --i)
cout << tmp[i] << "\t:" << counts[i] << "人\t"
<< std::fixed << std::setprecision(2) << rates[i] * 100 <<
"%" << endl;
}task2.cpp
#include "GradeCalc.hpp"
#include <iomanip>
void test() {
int n;
cout << "输入班级人数: ";
cin >> n;
GradeCalc c1("OOP", n);
cout << "录入成绩: " << endl;
c1.input();
cout << "输出成绩: " << endl;
c1.output();
cout << string(20, '*') + "课程成绩信息" + string(20, '*') << endl;
c1.info();
}
int main() {
test();
}
问题 1：派生类 GradeCalc 定义中，成绩存储在继承自 vector<int>的对象中（即 this 所指向的对象）。派生类方法 sort、min、max、average、output 通过迭代器访问成绩，例如 begin () 和 end () 获取迭代器范围来遍历成绩。input 方法通过 push_back 接口实现数据存入对象。
问题 2：代码 line68 分母 n 的功能是计算成绩的平均值，用于将总成绩除以班级人数得到平均成绩。去掉乘以 1.0 代码，重新编译、运行，结果会有影响。因为 accumulate 函数返回的是整数类型的累加和，如果不乘以 1.0，在计算平均值时会进行整数除法，导致结果不准确，乘以 1.0 是为了将结果转换为浮点数，以得到正确的平均值。
问题 3：从真实应用场景角度考虑，GradeCalc 类在设计及代码实现细节上可能存在以下不足：没有考虑成绩录入的合法性检查，例如录入负数或大于 100 的数；对于大规模数据的排序效率可能较低，可考虑使用更高效的排序算法；没有提供修改成绩的功能；在分数段统计中，对于边界值的处理可能不够灵活，例如 60 分既属于 [0, 60) 又属于 [60, 70) 的情况没有明确规定；没有考虑多门课程成绩管理的情况，可进一步扩展为可以管理多门课程成绩的类1
task3
GradeCalc.hpp
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>
#include <iomanip>
using std::vector;
using std::string;
using std::cin;
using std::cout;
using std::endl;
class GradeCalc {
public:
GradeCalc(const string &cname, int size);
void input(); //录入成绩
void output() const; //输出成绩
void sort(bool ascending = false); //排序(默认降序)
int min() const; //返回最低分
int max() const; //返回最高分
float average() const; //返回平均分
void info(); //输出课程成绩信息
private:
void compute(); //成绩统计
private:
string course_name; //课程名
int n; //课程人数
vector<int> grades;
vector<int> counts = vector<int>(5, 0); //保存各分数段人数([0, 60),[60,70),[70,80),[80,90),[90,100])
vector<double> rates = vector<double>(5, 0);//保存各分数段比例
};
GradeCalc::GradeCalc(const string &cname, int size) :
course_name(cname), n{size} {}
void GradeCalc::input() {
int grade;
for (int i = 0; i < n; ++i) {
cin >> grade;
grades.push_back(grade);
}
}
void GradeCalc::output() const {
for (int grade : grades)
cout << grade << " ";
cout << endl;
}
void GradeCalc::sort(bool ascending) {
if (ascending)
std::sort(grades.begin(), grades.end());
else
std::sort(grades.begin(), grades.end(), std::greater<int>());
}
int GradeCalc::min() const {
return *std::min_element(grades.begin(), grades.end());
}
int GradeCalc::max() const {
return *std::max_element(grades.begin(), grades.end());
}
float GradeCalc::average() const {
return std::accumulate(grades.begin(), grades.end(), 0) * 1.0 / n;
}
void GradeCalc::compute() {
for (int grade : grades) {
if (grade < 60)
counts.at(0)++;
else if (grade >= 60 && grade < 70)
counts.at(1)++;
else if (grade >= 70 && grade < 80)
counts.at(2)++;
else if (grade >= 80 && grade < 90)
counts.at(3)++;
else if (grade >= 90)
counts.at(4)++;
}
for (int i = 0; i < rates.size(); ++i)
rates.at(i) = counts.at(i) * 1.0 / n;
}
void GradeCalc::info() {
cout << "课程名称:\t" << course_name << endl;
cout << "排序后成绩:\t";
sort();
output();
cout << "最高分:\t" << max() << endl;
cout << "最低分:\t" << min() << endl;
cout << "平均分:\t" << std::fixed << std::setprecision(2) <<
average() << endl;
compute();//统计各分数段人数、比例
vector<string> tmp{"[0,60)", "[60,70)", "[70,80)", "[80,90)",
"[90,100]"};
for (int i = tmp.size() - 1; i >= 0; --i)
cout << tmp[i] << "\t:" << counts[i] << "人\t"
<< std::fixed << std::setprecision(2) << rates[i] * 100 <<
"%" << endl;
}task3.cpp
#include "GradeCalc.hpp"
#include <iomanip>
void test() {
int n;
cout << "输入班级人数: ";
cin >> n;
GradeCalc c1("OOP", n);
cout << "录入成绩: " << endl;
c1.input();
cout << "输出成绩: " << endl;
c1.output();
cout << string(20, '*') + "课程成绩信息" + string(20, '*') << endl;
c1.info();
}
int main() {
test();
}
问题 1：组合类 GradeCalc 定义中，成绩存储在私有数据成员 grades 中，它是一个 vector<int>类型的对象。组合类方法 sort、min、max、average、output 通过直接访问 grades 对象来获取每个成绩，例如在 sort 方法中直接使用 grades.begin () 和 grades.end () 获取迭代器范围进行排序操作。与实验任务 2 在代码写法细节上的差别在于，实验任务 2 是通过继承 vector<int>，利用继承关系访问成绩，而本任务是通过组合的方式，将 vector<int>作为类的一个成员来访问成绩。
问题 2：对比实验任务 2 和实验任务 3，对面向对象编程有以下新的理解和领悟：继承和组合是实现代码复用的两种重要方式。继承可以直接复用基类的功能，但可能会导致类之间的耦合度较高，如果基类发生变化，可能会影响到派生类。而组合则更加灵活，通过将其他类的对象作为成员，可以在不影响外部类的情况下修改内部成员类的实现，降低了类之间的耦合度。在实际编程中，应根据具体需求选择合适的方式来设计类，有时可能需要综合使用继承和组合来达到更好的设计效果。
task4
task4_1.cpp
#include <iostream>
#include <string>
#include <limits>
using namespace std;
void test1() {
string s1, s2;
cin >> s1 >> s2; // cin: 从输入流读取字符串, 碰到空白符(空格/回车/Tab)即结束
cout << "s1: " << s1 << endl;
cout << "s2: " << s2 << endl;
}
void test2() {
string s1, s2;
getline(cin, s1); // getline(): 从输入流中提取字符串,直到遇到换行符
getline(cin, s2);
cout << "s1: " << s1 << endl;
cout << "s2: " << s2 << endl;
}
void test3() {
string s1, s2;
getline(cin, s1, ' '); //从输入流中提取字符串,直到遇到指定分隔符
getline(cin, s2);
cout << "s1: " << s1 << endl;
cout << "s2: " << s2 << endl;
}
int main() {
cout << "测试1: 使用标准输入流对象cin输入字符串" << endl;
test1();
cout << endl;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "测试2: 使用函数getline()输入字符串" << endl;
test2();
cout << endl;
cout << "测试3: 使用函数getline()输入字符串, 指定字符串分隔符" << endl;
test3();
}
问题 1：去掉 task4_1.cpp 的 line35 后，重新编译、运行，当连续进行测试 1 和测试 2 时，会出现测试 2 直接跳过输入 s1 的情况。line35 的用途是忽略输入缓冲区中的剩余字符，直到遇到换行符 '\n'。在测试 1 中使用 cin 输入字符串时，输入完字符串后按下回车键，回车键会留在输入缓冲区中。如果不忽略这个回车键，在测试 2 中使用 getline () 函数读取字符串时，它会首先读取到这个回车键，导致直接跳过输入 s1 的步骤。
task2.cpp
#include <iostream>
#include <string>
#include <vector>
#include <limits>
using namespace std;
void output(const vector<string>&v) {
for (auto &s : v)
cout << s << endl;
}
void test() {
int n;
while (cout << "Enter n:", cin >> n) {
vector<string> v1;
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
v1.push_back(s);
}
cout << "output v1:" << endl;
output(v1);
cout << endl;
}
}
int main() {
cout << "测试:使用cin多组输入字符串" << endl;
test();
}
task4_3.cpp
#include <iostream>
#include <string>
#include <vector>
#include <limits>
using namespace std;
void output(const vector<string>&v) {
for (auto &s : v)
cout << s << endl;
}
void test() {
int n;
while (cout << "Enter n:", cin >> n) {
cin.ignore(numeric_limits<streamsize>::max(), '\n');
vector<string> v2;
for (int i = 0; i < n; ++i) {
string s;
getline(cin, s);
v2.push_back(s);
}
cout << "output v2:" << endl;
output(v2);
cout << endl;
}
}
int main() {
cout << "测试:使用函数getline()多组输入字符串" << endl;
test();
}问题 2：去掉 task4_3.cpp 的 line16 后，重新编译、运行，当输入数字后直接按回车键，然后输入字符串时，会出现程序直接结束的情况。line16 的用途是在每次输入数字后，忽略输入缓冲区中的剩余字符，直到遇到换行符 '\n'。这是因为在使用 cin 读取数字后，回车键会留在输入缓冲区中。如果不忽略这个回车键，当使用 getline () 函数读取字符串时，它会首先读取到这个回车键，导致 getline () 函数认为已经读取到了空行，从而直接结束本次循环，使程序无法正常输入字符串。
实验五
grm.hpp
template<typename T>
class GameResourceManager {
private:
T resource;
public:
GameResourceManager(T initialResource) : resource(initialResource) {}
T get() const {
return resource;
}
void update(T change) {
resource += change;
if (resource < 0) {
resource = 0;
}
}
};task5.cpp
#include "grm.hpp"
#include <iostream>
using std::cout;
using std::endl;
void test1() {
GameResourceManager<float> HP_manager(99.99);
cout << "当前生命值: " << HP_manager.get() << endl;
HP_manager.update(9.99);
cout << "增加9.99生命值后, 当前生命值: " << HP_manager.get() << endl;
HP_manager.update(-999.99);
cout << "减少999.99生命值后, 当前生命值: " << HP_manager.get() << endl;
}
void test2() {
GameResourceManager<int> Gold_manager(100);
cout << "当前金币数量: " << Gold_manager.get() << endl;
Gold_manager.update(50);
cout << "增加50个金币后, 当前金币数量: " << Gold_manager.get() << endl;
Gold_manager.update(-99);
cout << "减少99个金币后, 当前金币数量: " << Gold_manager.get() << endl;
}
int main() {
cout << "测试1: 用float类型对类模板GameResourceManager实例化" << endl;
test1();
cout << endl;
cout << "测试2: 用int类型对类模板GameResourceManager实例化" << endl;
test2();
}
task6
info.hpp
#include <string>
class Info {
private:
std::string nickname;
std::string contact;
std::string city;
int n;
public:
Info(const std::string &nick, const std::string &cont, const std::string &cty, int num) :
nickname(nick), contact(cont), city(cty), n(num) {}
void display() const {
std::cout << "昵称: " << nickname << std::endl;
std::cout << "联系方式: " << contact << std::endl;
std::cout << "所在城市: " << city << std::endl;
std::cout << "预定人数: " << n << std::endl;
}
};task6.cpp
#include "info.hpp"
#include <iostream>
#include <vector>
#include <limits>
const int capacity = 100;
int main() {
std::vector<Info> audience_list;
int total = 0;
while (true) {
std::string nick, cont, city;
int num;
std::cout << "录入用户预约信息: " << std::endl;
std::cout << "昵称 联系方式(邮箱/手机号) 所在城市 预定参加人数" << std::endl;
if (!(std::cin >> nick >> cont >> city >> num)) {
break;
}
if (total + num > capacity) {
std::cout << "对不起,只剩 " << capacity - total << " 个位置. 输入u,更新(update)预定信息，输入q,退出预定" << std::endl;
std::string choice;
std::cin >> choice;
if (choice == "u") {
std::cout << "请重新输入预定信息: " << std::endl;
continue;
} else if (choice == "q") {
break;
}
}
audience_list.push_back(Info(nick, cont, city, num));
total += num;
if (total >= capacity) {
std::cout << "截至目前,一共有 " << total << " 位听众预约。预约听众信息如下: " << std::endl;
for (const auto &info : audience_list) {
info.display();
}
break;
}
}
return 0;
}
task7
date.h
1 #pragma once
2 #ifndef DATE H
3 #define DATE H
4 class Date {
5 private:
6 int year;
7 int month;
8 int day;
9 int totalDays;
10 public:
11 Date(int year, int month, int day);
12 int getYear()const { return year; }
13 int getMonth()const { return month; }
14 int getDay()const { return day; }
15 int getMaxDay()const;
16 bool isLeapYear()const {
17 return year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
18 }
19 void show()const;
20 int distance(const Date& date)const {
21 return totalDays - date.totalDays;
22 }
23 };
24 #endif// DATE H
date.hdate.cpp
1 #include"date.h"
2 #include<iostream>
3 #include<cstdlib>
4 using namespace std;
5 namespace {
6 const int DAYS_BEFORE_MONTH[] = { 0,31,59,90,120,151,181,212,243,273,304,334,365 };
7 }
8 Date::Date(int year, int month, int day) :year{ year }, month{ month }, day{ day } {
9 if (day <= 0 || day > getMaxDay()) {
10 cout << "Invalid date:";
11 show();
12 cout << endl;
13 exit(1);
14 }
15 int years = year - 1;
16 totalDays = years * 365 + years / 4 - years / 100 + years / 400 + DAYS_BEFORE_MONTH[month - 1] + day;
17 if (isLeapYear() && month > 2)totalDays++;
18 }
19 int Date::getMaxDay()const {
20 if (isLeapYear() && month == 2)
21 return 29;
22 else return DAYS_BEFORE_MONTH[month] - DAYS_BEFORE_MONTH[month - 1];
23 }
24
25 void Date::show()const {
26 cout << getYear() << "-" << getMonth() << "-" << getDay();
27 }
date.cppaccumulator.h
1 #pragma once
2 #ifndef ACCUMULATOR H
3 #define ACCUMULATOR H
4 #include"date.h"
5 class Accumulator {
6 private:
7 Date lastDate;
8 double value;
9 double sum;
10 public:
11 Accumulator(const Date& date, double value) :lastDate(date), value(value), sum{ 0 } {
12 }
13
14 double getSum(const Date& date)const {
15 return sum + value * date.distance(lastDate);
16 }
17
18 void change(const Date& date, double value) {
19 sum = getSum(date);
20 lastDate = date; this->value = value;
21 }
22
23 void reset(const Date& date, double value) {
24 lastDate = date; this->value = value; sum = 0;
25 }
26 };
27 #endif//ACCUMULATOR H
accumulator.haccount.h
1 #pragma once
2 #ifndef ACCOUNT H
3 #define ACCOUNT H
4 #include"date.h"
5 #include"accumulator.h"
6 #include<string>
7 class Account {
8 private:
9 std::string id;
10 double balance;
11 static double total;
12 protected:
13 Account(const Date& date, const std::string& id);
14 void record(const Date& date, double amount, const std::string& desc);
15 void error(const std::string& msg)const;
16 public:
17 const std::string& getId()const { return id; }
18 double getBalance()const { return balance; }
19 static double getTotal() { return total; }
20
21 void show()const;
22 };
23 class SavingsAccount :public Account {
24 private:
25 Accumulator acc;
26 double rate;
27 public:
28 SavingsAccount(const Date& date, const std::string& id, double rate);
29 double getRate()const { return rate; }
30
31 void deposit(const Date& date, double amount, const std::string& desc);
32 void withdraw(const Date& date, double amount, const std::string& desc);
33 void settle(const Date& date);
34 };
35 class CreditAccount :public Account {
36 private:
37 Accumulator acc;
38 double credit;
39 double rate;
40 double fee;
41 double getDebt()const {
42 double balance = getBalance();
43 return (balance < 0 ? balance : 0);
44 }
45 public:
46 CreditAccount(const Date& date, const std::string& id, double credit, double rate, double fee);
47 double getCredit()const { return credit; }
48 double getRate()const { return rate;}
49 double getAvailableCredit()const {
50 if (getBalance() < 0)
51 return credit + getBalance();
52 else
53 return credit;
54 }
55 void deposit(const Date& date, double amount, const std::string& desc);
56 void withdraw(const Date& date, double amount, const std::string& desc);
57 void settle(const Date& date);
58 void show()const;
59 };
60 #endif//ACCOUNT H
account.h:account.cpp
1 #include"account.h"
2 #include<cmath>
3 #include<iostream>
4 using namespace std;
5 double Account::total = 0;
6
7 Account::Account(const Date& date, const string& id) :id{ id }, balance{ 0 } {
8 date.show(); cout << "\t#" << id << "created" << endl;
9 }
10
11
12 void Account::record(const Date& date, double amount, const string& desc) {
13 amount = floor(amount * 100 + 0.5) / 100;
14 balance += amount;
15 total += amount;
16 date.show();
17 cout << "\t#" << id << "\t" << amount << "\t" << balance << "\t" << desc << endl;
18 }
19
20 void Account::show()const { cout << id << "\tBalance:" << balance; }
21 void Account::error(const string& msg)const {
22 cout << "Error(#" << id << "):" << msg << endl;
23 }
24
25 SavingsAccount::SavingsAccount(const Date&date,const string&id,double rate):Account(date,id),rate(rate), acc(date,0){}
26
27 void SavingsAccount::deposit(const Date& date, double amount, const string& desc) {
28 record(date, amount, desc);
29 acc.change(date, getBalance());
30 }
31
32 void SavingsAccount::withdraw(const Date& date, double amount, const string& desc) {
33 if (amount > getBalance()) {
34 error("not enough money");
35 }
36 else {
37 record(date, -amount, desc);
38 acc.change(date, getBalance());
39 }
40 }
41
42 void SavingsAccount::settle(const Date& date) {
43 double interest = acc.getSum(date) * rate / date.distance(Date(date.getYear() - 1, 1, 1));
44 if (interest != 0)record(date, interest, "interest");
45 acc.reset(date, getBalance());
46 }
47
48 CreditAccount::CreditAccount(const Date&date,const string&id,double credit,double rate,double fee):Account(date,id),credit(credit),rate(rate),fee(fee),acc(date,0){}
49
50 void CreditAccount::deposit(const Date& date, double amount, const string& desc) {
51 record(date, amount, desc);
52 acc.change(date, getDebt());
53 }
54
55 void CreditAccount::withdraw(const Date& date, double amount, const string& desc) {
56 if (amount - getBalance() > credit) {
57 error("not enough credit");
58 }
59 else {
60 record(date, -amount, desc);
61 acc.change(date, getDebt());
62 }
63 }
64
65 void CreditAccount::settle(const Date& date) {
66 double interest = acc.getSum(date) * rate;
67 if (interest != 0)record(date, interest, "interest");
68 if (date.getMonth() == 1)
69 record(date, -fee, "annual fee");
70 acc.reset(date, getDebt());
71 }
72
73 void CreditAccount::show()const {
74 Account::show();
75 cout << "\tAvailable credit:" << getAvailableCredit();
76 }
account.cpptask7_10.cpp
1 #include"account.h"
2 #include<iostream>
3
4 using namespace std;
5
6 int main() {
7 Date date(2008, 11, 1);
8 SavingsAccount sa1(date, "S3755217", 0.015);
9 SavingsAccount sa2(date, "02342342", 0.015);
10 CreditAccount ca(date, "C5392394", 10000, 0.0005, 50);
11
12 sa1.deposit(Date(2008, 11, 5), 5000, "salary");
13 ca.withdraw(Date(2008, 11, 15), 2000, "buy a cell");
14 sa2.deposit(Date(2008, 11, 25), 10000, "sell stock 0323");
15
16 ca.settle(Date(2008, 12, 1));
17
18 ca.deposit(Date(2008, 12, 1), 2016, "repay the credit");
19 sa1.deposit(Date(2008, 12, 5), 5500, "salary");
20
21 sa1.settle(Date(2009, 1, 1));
22 sa2.settle(Date(2009, 1, 1));
23 ca.settle(Date(2009, 1, 1));
24
25 cout << endl;
26 sa1.show(); cout << endl;
27 sa2.show(); cout << endl;
28 ca.show(); cout << endl;
29 cout << "Total:" << Account::getTotal() << endl;
30 return 0;
31 }
task7_10.cpp
使用了类的继承，通过基类account继承两个派生类SavingAccounts和CreditAccounts,模块抽象更清晰。
代码复杂度增加。
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