Algorithm to add or subtract days from a date?

问题:

I'm trying to write a Date class in an attempt to learn C++.

I'm trying to find an algorithm to add or subtract days to a date, where Day starts from 1 and Month starts from 1. It's proving to be very complex, and google doesn't turn up much,

Does anyone know of an algorithm which does this?

回答1:

The easiest way is to actually write two functions, one which converts the day to a number of days from a given start date, then another which converts back to a date. Once the date is expressed as a number of days, it's trivial to add or subtract to it.

You can find the algorithms here: http://alcor.concordia.ca/~gpkatch/gdate-algorithm.html

回答2:

You don't really need an algorithm as such (at least not something worthy of the name), the standard library can do most of the heavy lifting; calender calculations are notoriously tricky. So long as you don't need dates earlier than 1900, then:

#include <ctime>

// Adjust date by a number of days +/-
void DatePlusDays( struct tm* date, int days )
{
    const time_t ONE_DAY = 24 * 60 * 60 ;

    // Seconds since start of epoch
    time_t date_seconds = mktime( date ) + (days * ONE_DAY) ;

    // Update caller's date
    // Use localtime because mktime converts to UTC so may change date
    *date = *localtime( &date_seconds ) ; ;
}

Example usage:

#include <iostream>

int main()
{
    struct tm date = { 0, 0, 12 } ;  // nominal time midday (arbitrary).
    int year = 2010 ;
    int month = 2 ;  // February
    int day = 26 ;   // 26th

    // Set up the date structure
    date.tm_year = year - 1900 ;
    date.tm_mon = month - 1 ;  // note: zero indexed
    date.tm_mday = day ;       // note: not zero indexed

    // Date, less 100 days
    DatePlusDays( &date, -100 ) ; 

    // Show time/date using default formatting
    std::cout << asctime( &date ) << std::endl ;
}

回答3:

I'm assuming this is for some kind of an exercise, otherwise you would use a time class that's already provided to you.

You could store your time as the number of milliseconds since a certain date. And then you can add the appropriate value and convert from that to the date upon calling the accessors of your class.

回答4:

Here's a sketch of a very simple approach. For simplicity of ideas I will assume that d, the number of days to add, is positive. It is easy to extend the below to cases where d is negative.

Either d is less than 365 or d is greater than or equal to 365.

If d is less than 365:

m = 1;
while(d > numberOfDaysInMonth(m, y)) {
    d -= numberOfDaysInMonth(m, y);
    m++;
}
return date with year = y, month = m, day = d;

If d is greater than 365:

while(d >= 365) {
    d -= 365;
    if(isLeapYear(y)) {
        d -= 1;
    }
    y++;
}
// now use the case where d is less than 365

Alternatively, you could express the date in, say, Julian form and then merely add to the Julian form and conver to ymd format.

回答5:

One approach is to map the date to the Julian number of the date, do your integer operations and then transform back.

You will find plenty of resources for the julian functions.

回答6:

Try this function. It correctly calculates additions or subtractions. dateTime argument must be in UTC format.

tm* dateTimeAdd(const tm* const dateTime, const int& days, const int& hours, const int& mins, const int& secs) {
    tm* newTime = new tm;
    memcpy(newTime, dateTime, sizeof(tm));

    newTime->tm_mday += days;
    newTime->tm_hour += hours;
    newTime->tm_min += mins;
    newTime->tm_sec += secs;        

    time_t nt_seconds = mktime(newTime) - timezone;
    delete newTime;

    return gmtime(&nt_seconds);
}

And there are example of using:

time_t t = time(NULL);
tm* utc = gmtime(&t);
tm* newUtc = dateTimeAdd(utc, -5, 0, 0, 0); //subtract 5 days

回答7:

I would suggest writing first a routine which converts year-month-day into a number of days since fixed date, say, since 1.01.01. And a symmetric routine which would convert it back.

Don't forget to process leap years correctly!

Having those two, your task would be trivial.

回答8:

I know this is a very old question but it's an interesting and some common one when it comes to working with dates and times. So I thought of sharing some code which calculates the new date without using any inbuilt time functionality in C++.

#include <iostream>
#include <string>

using namespace std;

class Date {
public:
    Date(size_t year, size_t month, size_t day):m_year(year), m_month(month), m_day(day) {}
    ~Date() {}

    // Add specified number of days to date
    Date operator + (size_t days) const;

    // Subtract specified number of days from date
    Date operator - (size_t days) const;

    size_t Year()  { return m_year; }
    size_t Month() { return m_month; }
    size_t Day()   { return m_day; }

    string DateStr();
private:
    // Leap year check 
    inline bool LeapYear(int year) const
        { return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0); }

    // Holds all max days in a general year
    static const int MaxDayInMonth[13];

    // Private members
    size_t m_year;
    size_t m_month;
    size_t m_day;   
};

// Define MaxDayInMonth
const int Date::MaxDayInMonth[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

//===========================================================================================
/// Add specified number of days to date
Date Date::operator + (size_t days) const {
    // Maximum days in the month
    int nMaxDays(MaxDayInMonth[m_month] + (m_month == 2 && LeapYear(m_year) ? 1 : 0));

    // Initialize the Year, Month, Days
    int nYear(m_year);
    int nMonth(m_month);
    int nDays(m_day + days);

    // Iterate till it becomes a valid day of a month
    while (nDays > nMaxDays) {
        // Subtract the max number of days of current month
        nDays -= nMaxDays;

        // Advance to next month
        ++nMonth;

        // Falls on to next year?
        if (nMonth > 12) {
            nMonth = 1; // January
            ++nYear;    // Next year
        }

        // Update the max days of the new month
        nMaxDays = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);
    }

    // Construct date
    return Date(nYear, nMonth, nDays);
}

//===========================================================================================
/// Subtract specified number of days from date
Date Date::operator - (size_t days) const {
    // Falls within the same month?
    if (0 < (m_day - days)) {
        return Date(m_year, m_month, m_day - days);
    }

    // Start from this year
    int nYear(m_year);

    // Start from specified days and go back to first day of this month
    int nDays(days);
    nDays -= m_day;

    // Start from previous month and check if it falls on to previous year
    int nMonth(m_month - 1);
    if (nMonth < 1) {
        nMonth = 12; // December
        --nYear;     // Previous year
    }

    // Maximum days in the current month
    int nDaysInMonth = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);

    // Iterate till it becomes a valid day of a month
    while (nDays >= 0) {
        // Subtract the max number of days of current month
        nDays -= nDaysInMonth;

        // Falls on to previous month?
        if (nDays > 0) {
            // Go to previous month
            --nMonth;

            // Falls on to previous year?
            if (nMonth < 1) {
                nMonth = 12; // December
                --nYear;     // Previous year
            }
        }

        // Update the max days of the new month
        nDaysInMonth = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);
    }

    // Construct date
    return Date(nYear, nMonth, (0 < nDays ? nDays : -nDays));
}

//===========================================================================================
/// Get the date string in yyyy/mm/dd format
string Date::DateStr() {
    return to_string(m_year) 
        + string("/")
        + string(m_month < 10 ? string("0") + to_string(m_month) : to_string(m_month))
        + string("/")
        + string(m_day < 10 ? string("0") + to_string(m_day) : to_string(m_day)); 
}


int main() {
    // Add n days to a date
    cout << Date(2017, 6, 25).DateStr() << " + 10 days = "
         << (Date(2017, 6, 25) /* Given Date */ + 10 /* Days to add */).DateStr() << endl;

    // Subtract n days from a date
    cout << Date(2017, 6, 25).DateStr() << " - 10 days = "
         << (Date(2017, 6, 25) /* Given Date */ - 10 /* Days to subract */).DateStr() << endl;

    return 0;
}

Output
2017/06/25 + 10 days = 2017/07/05
2017/06/25 - 10 days = 2017/06/15