Date class and time class definitions (operator overload application)

Date class and time class definitions （ Operator overload application ）

Date class definition （ Operator overload application ）：

class Data     // At first, it was only used to indicate the date of publication , Later, it was found that both borrowing and return dates were available 
{
    
    int year;
    int month;
    int day;
public:
    Data(){
    };
    Data(int year1,int month1,int day1):year(year1),month(month1),day(day1){
    };
    friend ostream&operator<<(ostream&os,const Data&d);
    friend istream&operator>>(istream&is,Data&d);
    bool operator<(const Data &d)const    // Date comparison function 
    {
    return year!=d.year?year<d.year:month!=d.month?month<d.month:day<d.day;}
    int operator-(Data d)  // The operation of the date , You can directly subtract two months by month , You can also operate a little more accurately as follows （ It can also be used to judge the normal and leap years for more accurate operation ）
	{
    
	    if(year<d.year||(year==d.year&&month<d.month)||(year==d.year&&month==d.month&&day<d.day))
            return -1;  // There is no way to subtract the date （ When is the later date ）
        int m[13] = {
    0,31,28,31,30,31,30,31,31,30,31,30,31};  // Press the subscript （ Corresponding to the month ） A filling , The first number is not counted , For February of each year, press 28 Tianji （ Do not consider the leap year ）
	    int count=0;
	    int a=d.year,b=d.month,c=d.day;
	    while(1)
	    {
    
            if(this->year==a&&this->month==b&&this->day==c) break;
	        if(c==m[b]){
    
                c=1;
                if(b==12)
                 {
    
                     b=1;
                     a++;
                 }
                 else b++;
	        }
	        else c++;
	        count++;
        }
	        return count;
	}
	Data operator+ (int x)
	{
    
        int m[13] = {
    0,31,28,31,30,31,30,31,31,30,31,30,31};  // Press the subscript （ Corresponding to the month ） A filling , The first number is not counted , For February of each year, press 28 Tianji （ Do not consider the leap year ）
        while(x--)
        {
    
            if(day==m[month])
            {
    
                day=1;
                if(month==12)
                {
    
                    year++;
                    month=1;
                }
                else month++;
            }
            else day++;
        }
        return *this;
	}
};

istream&operator>>(istream&is,Data&d)    // First reload the input and output of the publication date 
{
    
    while(1)
    {
    
        is>>d.year>>d.month>>d.day;
        if(d.year<1900||d.year>2019||d.month<1||d.month>12||d.day<1||d.day>31); // Here, it can be optimized to judge leap years or ordinary years, and then judge 
        else break;
    }
    return is;
}
ostream&operator<<(ostream&os,const Data&d)
{
    
    //os<<"year:"<<d.year<<" "<<"month:"<<d.month<<" "<<"day:"<<d.day;
    // In order to unify input and output , Change to the following output form 
    os<<d.year<<" "<<d.month<<" "<<d.day;
    return os;
}

Time class definition （ Operator overload application ）：

class Time
{
    
    int year,month,day,hour,min;
    string s;
public:
    Time() {
    };
    ~Time() {
    };
    Time(int year1,int month1 ,int day1,int hour1,int min1):year(year1),month(month1),day(day1),hour(hour1),min(min1) {
    };
    friend istream& operator>>(istream&,Time&t);
    friend ostream& operator<< (ostream&,const Time&t);
    int operator- (Time t)  // Overload operator "-" Define... Within a class , Out of class definitions due to year,month Waiting is private , Need to use get Function fetch data , There is no need to 
    {
       // Just define it directly , There is no need to use references , At first, I felt the same , Using references , Later, it was found that the error could not be adjusted , It took a long time to find out , After using the reference, the value after the minus sign will become the value before the minus sign , The two values are the same 
        if((year<t.year)||(year==t.year&&month<t.month)||(year==t.year&&month==t.month&&day<t.day)||(year==t.year&&month==t.month&&day==t.day&&hour<t.hour)||year==t.year&&month==t.month&&day==t.day&&hour==t.hour&&min<t.min)
        return -1;
        int countyear=0;
        int countmonth=0;
        int countday=0;
        int counthour=0;
        int countmin=0;
        int m[13] = {
    0,31,28,31,30,31,30,31,31,30,31,30,31};  // Press the subscript （ Corresponding to the month ） A filling , The first number is not counted , For February of each year, press 28 Tianji （ Do not consider the leap year ）
        while(1)
	    {
    
            if(this->year==t.year&&this->month==t.month&&this->day==t.day&&hour==t.hour&&min==t.min) break;
	        if(t.min==60)
	        {
    
	            t.hour++;
	            counthour++;
	            t.min=0;
                if(t.hour==24)
                {
    
                    t.day++;
                    countday++;
                    t.hour=0;
                    if(t.day>m[t.month])  // The number of days at the end of the above date exists , If write "==", Then skip directly , Unrealistic 
                    {
    
                        t.month++;
                        countmonth++;
                        t.day=1;
                        if(t.month>12)    //12 The moon exists 
                        {
    
                            t.year++;
                            countyear++;
                            t.month=1;
                        }
                    }
                }
	        }
	        else {
    t.min++;countmin++;}
        }
        //cout<<countyear<<" "<<countmonth<<" "<<countday<<" "<<counthour<<" "<<countmin<<endl;
        return countmin;
    }
    //Time operator+ (int i); // utilize "-" Operation can replace "+"
    // For the comparison of time , I didn't plan to use it at first , However, the later login and the view of the user's travel ticket need to be used 
    bool operator< (const Time&t)const
    {
    
        return year!=t.year?year<t.year:month!=t.month?month<t.month:day!=t.day?day<t.day:hour!=t.hour?hour<t.hour:min<t.min;
    }
    bool operator== (const Time&t)const
    {
    
        return year==t.year&&month==t.month&&day==t.day&&hour==t.hour&&min==t.min;
    }
};

istream&operator>>(istream&in,Time&t)
{
    
    while(1)  // It's good or bad to use circulation , When the program is running, the user can input the time again , But when the time in the file is wrong , The program will loop all the time 
    {
    
        //in>>t.year>>t.s>>t.month>>t.s>>t.day>>t.s>>t.hour>>t.s>>t.min;
        in>>t.year>>t.month>>t.day>>t.hour>>t.min;
        if(t.year>2000&&t.year<=2019&&t.month>=1&&t.month<=12&&t.day>=1&&t.day<=31&&t.hour>=0&&t.hour<=23&&t.min>=0&&t.min<60)   // Limit time , It should also be noted that outdated tickets cannot be bought （ Query is also not available , This can be requested in login time ）
        break;
        else cout<<"Time error, Please try again ："<<endl;
    }
    return in;
}

ostream&operator<<(ostream&out,const Time&t)
{
    
    out<<t.year<<" "<<t.month<<" "<<t.day<<" "<<t.hour<<" "<<t.min;
    return out;
}