Arithmetics on calendar dates in C or C++ (add N days to given date)

Question

I have been given a date, Which I am taking as an input like (day, month, year): 12, 03, 87.

Now I need to find out the date after n days.<

Answer 1

Can be implemented using C++ operators and in a quite OOP way by representing date as a class.

#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;

    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);
}

/// 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;

    return 0;
}

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

Answer 2

Here's what the solution looks like using algorithms published here by Howard Hinnant.

int main() {
    int day_count = days_from_civil(1980, 5, 2);
    auto [year, month, day] = civil_from_days(day_count + 40);
    printf("%04i-%02i-%02-i\n", (int)year, (int)month, (int)day);
}

New date functionality has been approved for inclusion in C++20, though with a different API. A C++20 solution will probably look something like:

#include <chrono>

int main() {
    using namespace std::chrono;
    sys_days in_days = year_month_day{1980y, may, 2d};
    std::cout << year_month_day(in_days + days(40)) << '\n';
}

Answer 3

the easiest trick is to use time_t type and corresponding functions.

mktime will convert tm structure to time_t. which is an integer value counting the seconds starting 01-Jan-1970.

After you have a time_t value just add the seconds count you need (86400 per day).

To convert back, use gmtime or localtime

Answer 4

This code will print the date coming after 10 days. Change the value to N for a user-defined number.

#include< iostream.h>

#include< conio.h>

struct date{int d,m,y;};

void main()

    {

    date d1;

    void later (date);

    cout<<"ENTER A VALID DATE (dd/mm/yy)";

    cin>>d1.d>>d1.m>>d1.y;

    later(d1);

    getch();

    }

    void later(date d1)

    {

    int mdays[12]={31,28,31,30,31,30,31,31,30,31,30,31};

    if(((d1.y%4==0) && (d1.y%100!=0))||(d1.y%400==0))
    mdays[1]=29;
    d1.d=d1.d+10;

    if(d1.d>mdays[d1.m-1])

    {

    d1.d=d1.d-mdays[d1.m-1];

    d1.m++;

    if(d1.m>12)

    {d1.m=1;

    d1.y++;

    }

    }
    cout<<"DATE AFTER TEN DAYS IS "<<d1.d<<"\t"<<d1.m<<"\t"<<d1.y;

    getch();

    }

Answer 5

Can you please tell me any good logic which works faster and have less complexity.

If this exact thing is indeed a performance critical part of your application, you're likely doing something wrong. For the sake of clarity and correctness, you should stick to the existing solutions. Select the one that is most appropriate to your development environment.

---

The C approach:

#include <stdio.h>
#include <time.h>

int main()
{        
    /* initialize */
    int y=1980, m=2, d=5;    
    struct tm t = { .tm_year=y-1900, .tm_mon=m-1, .tm_mday=d };
    /* modify */
    t.tm_mday += 40;
    mktime(&t);
    /* show result */
    printf("%s", asctime(&t)); /* prints: Sun Mar 16 00:00:00 1980 */
    return 0;
}

The C++ without using Boost approach:

#include <ctime>
#include <iostream>

int main()
{        
    // initialize
    int y=1980, m=2, d=5;
    std::tm t = {};
    t.tm_year = y-1900;
    t.tm_mon  = m-1;
    t.tm_mday = d;
    // modify
    t.tm_mday += 40;
    std::mktime(&t);
    // show result
    std::cout << std::asctime(&t); // prints: Sun Mar 16 00:00:00 1980
}

The Boost.Date_Time approach:

#include <boost/date_time/posix_time/posix_time.hpp>
#include <iostream>

int main()
{
    using namespace boost::gregorian;
    // initialize
    date d(1980,2,5);
    // modify
    d += days(40);
    // show result
    std::cout << d << '\n'; // prints: 1980-Mar-16
}

Answer 6

Just add to a time_t object for how many days you want.

#define SECOND              1
#define MINUTE              60 * SECOND
#define HOUR                60 * MINUTE
#define DAY                 24 * HOUR





time_t curTime;
time_t futureTime;

time( & curTime );

futureTime = curTime + (5 * DAY);

struct tm * futureDate = gmtime(&futureTime);
std::cout<<"5 days from now will be: "<<futureDate->tm_mday<<"/"<<futureDate->tm_mon+1<<"/"<<futureDate->tm_year+1900<<std::endl;