c++作业


输入输出流与文件操作

#include<iostream>
#include<fstream>
#include<cstdlib>
#include<string>

using namespace std;

//职工
struct Person
{
    int m_Id;
    string m_Name;
    int m_Age;
    int m_Wage;

    Person(int i, string n, int a, int w): m_Id(i), m_Name(n), m_Age(a), m_Wage(w) { }
    void PrintInfo()
    {
        cout << "===================" << endl;
        cout << "号码:\t" << this -> m_Id << endl;
        cout << "名字:\t" << this -> m_Name << endl;
        cout << "年龄:\t" << this -> m_Age << endl;
        cout << "工资:\t" << this -> m_Wage << endl;
        cout << "===================" << endl;
    }
};

Person *p[3] =
{
    new Person(101, "小黄", 23, 3100),
    new Person(102, "中黄", 41, 6600),
    new Person(105, "大黄", 55, 7000),

};

int main()
{
    //对5组数据按ID进行排序
    int i, j;
    Person *temp;
    for(i = 0; i < 3; i++)
    {
        for(j = i + 1; j < 3; j++)
        {
            if(p[i] -> m_Id > p[j] -> m_Id)
            {
                temp = p[i];
                p[i] = p[j];
                p[j] = temp;
            }
        }
    }

    //将排好序的数据存入到文件f.dat
    ofstream outfile1("f.dat", ios :: out);
    if(!outfile1)
    {
        cerr << "f.dat 打开错误" << endl;
        exit(1);
    }
    for(i = 0; i < 3; i++)
    {
        outfile1 << p[i] -> m_Id << " "
                 << p[i] -> m_Name << " "
                 << p[i] -> m_Age << " "
                 << p[i] -> m_Wage << endl;
    }
    outfile1.close();

    //再写入两个职工信息
    //示例输入 108 Titus 40 4000 109 Tulius 35 5000
    cout << "请输入2个职工信息:";
    ofstream outfile2("f.dat", ios :: out | ios :: app);
    if(!outfile2)
    {
        cerr << "f.dat 打开错误" << endl;
        exit(1);
    }
    temp = new Person(0, "", 0, 0);
    cin >> temp -> m_Id >> temp -> m_Name >> temp -> m_Age >> temp -> m_Wage;
    outfile2 << temp -> m_Id << " "
             << temp -> m_Name << " "
             << temp -> m_Age << " "
             << temp -> m_Wage << endl;
    cin >> temp -> m_Id >> temp -> m_Name >> temp -> m_Age >> temp -> m_Wage;
    outfile2 << temp -> m_Id << " "
             << temp -> m_Name << " "
             << temp -> m_Age << " "
             << temp -> m_Wage << endl;
    outfile2.close();

    //输出文件中的全部职工数据
    ifstream infile1("f.dat", ios :: in);
    if(!infile1)
    {
        cerr << "f.dat 打开错误" << endl;
        exit(1);
    }
    for(i = 0; i < 5; i++)
    {
        infile1 >> temp -> m_Id >> temp -> m_Name >> temp -> m_Age >> temp -> m_Wage;
        temp -> PrintInfo();
    }
    infile1.close();

    //输入一个ID检测是否存在
    int id;
    cout << "请输入一个号码:";
    cin >> id;
    bool isFound = false;
    ifstream infile2("f.dat", ios :: in);
    if(!infile2)
    {
        cerr << "f.dat 打开错误" << endl;
        exit(1);
    }
    for(i = 0; i < 5; i++)
    {
        infile2 >> temp -> m_Id >> temp -> m_Name >> temp -> m_Age >> temp -> m_Wage;
        if(temp -> m_Id == id)
        {
            temp -> PrintInfo();
            isFound = true;
            break;
        }
    }
    if(!isFound)
    {
        cerr << "没有找到这个人" << endl;
    }
    infile2.close();

    return 0;
}

类和对象的进阶

  1. 定义一个复数类,重载运算符“+”、“-”、“”、“/”,使这四个运算符能应用于复数的加减乘除运算。 复数加法:m=a+bi ,n=c+di; m+n=(a+c)+(b+d)i; 复数减法:m=a+bi ,n=c+di; m-n=(a-c)+(b-d)i; 复数乘法:m=a+bi ,n=c+di; mn=(ac-bd)+(bc+ad)i;
    复数除法:m=a+bi ,n=c+di; m/n=((ac+bd)+(bc-ad)i)/(c2+d2)。
  2. 参考教材例题11.1,11.2,11.3,11.5,分别采用公有继承,私有继承和保护继承的方式,在程序运行中,输入num,name,sex,age,addr的值,程序应输出以上五个数据的数值;
  3. 3.将教材例9.13中的Time类声明为Date的友元类,输出年月日和时分秒。

第一题

#include <iostream>
#include<cstdio>
using namespace std;
class Complex
{
public:
    Complex(double x = 0, double y = 0) :real(x), imag(y) {}
    friend Complex operator * (Complex &m, Complex &n);
    friend Complex operator / (Complex &m, Complex &n);
    Complex operator + (Complex &n);
    Complex operator - (Complex &n);
    void display();
private:
    double real;
    double imag;
};
Complex operator * (Complex &m, Complex &n)
{
    return Complex(m.real*n.real - m.imag*n.imag, m.real*n.imag + m.imag*n.real);
}
Complex operator / (Complex &m, Complex &n)
{

    Complex c;
    c.real = (m.real*n.real + m.imag*n.imag) / (n.real*n.real + n.imag*n.imag);
    c.imag = (m.imag*n.real - m.real*n.imag) / (n.real*n.real + n.imag*n.imag);
    return c;
}
Complex  Complex ::operator + (Complex &n)
{
    return Complex(real + n.real, imag + n.imag);
}
Complex  Complex ::operator - (Complex &n)
{
    return Complex(real - n.real, imag - n.imag);
}
void Complex::display()
{
    cout << "<" << real << "," << imag << "i>" << endl;
}
int main()
{
    int a,bi,c,di;
    cout<<"请输入m的实部和虚部"<<endl;
    cin>>a>>bi;
    cout<<"请输入n的实部和虚部"<<endl;
    cin>>c>>di;
    Complex m(a, bi), n(c, di), c3;
    cout<<"为"<<endl;
    m.display();
    cout<<"n为"<<endl;
    n.display();
    cout<<"计算结果为"<<endl;
    c3 = m + n;
    c3.display();
    c3 = m - n;
    c3.display();
    c3 = m * n;
    c3.display();
    c3 = m / n;
    c3.display();
    return 0;
}

第二题

#include <iostream>
using namespace std;
class Student
{
public:
    void get_value()
    {
        cin>>num>>name>>sex;
    }
    void display( )
    {
        cout<<"num: "<<num<<endl;
        cout<<"name: "<<name<<endl;
        cout<<"sex: "<<sex<<endl;
    }
private :
    int num;
    string name;
    char sex;
};
class Student1:public Student
{
public:
    void get_value1()
    {
        get_value();
        cin>>age>>addr;
    }
    void display1()
    {
        display();
        cout<<"age: "<<age<<endl;
        cout<<"address: "<<addr<<endl;
    }
private:
    int age;
    string addr;
};
class Student2:private Student
{
public:
    void get_value1()
    {
        get_value();
        cin>>age>>addr;
    }
    void display1()
    {
        display();
        cout<<"age: "<<age<<endl;
        cout<<"address: "<<addr<<endl;
    }
private:
    int age;
    string addr;
};
class Student3:protected Student
{
public:
    void get_value1()
    {
        get_value();
        cin>>age>>addr;
    }
    void display1()
    {
        display();
        cout<<"age: "<<age<<endl;
        cout<<"address: "<<addr<<endl;
    }
private:
    int age;
    string addr;
};
int main()
{
    Student1 stud1;
    stud1.get_value1();
    stud1.display1();
    Student2 stud2;
    stud2.get_value1();
    stud2.display1();
    Student3 stud3;
    stud3.get_value1();
    stud3.display1();
    return 0;
}

第三题

#include <iostream>
using namespace std;
class Date;
class Time
{
public:
    Time(int h, int m, int s): hour(h), minute(m), sec(s) {};
    void display(Date &);
private:
    int hour;
    int minute;
    int sec;
};
class Date
{
public:
    friend Time;
    Date(int m, int d, int y): month(m), day(d), year(y) {};
private:
    int month;
    int day;
    int year;
};
void Time::display(Date &d)
{
    cout<<d.month<<"/"<<d.day<<"/"<<d.year<<endl;
    cout<<hour<<":"<<minute<<":"<<sec<<endl;
}
int main()
{
    Date d1(01, 01, 2019);
    Time t1(15, 57, 56);
    t1.display(d1);
    return 0;
}

文章作者: Czq
版权声明: 本博客所有文章除特別声明外,均采用 CC BY 4.0 许可协议。转载请注明来源 !
评论
  目录