easy way to add 1 month to a time_t in C/C++

Question

I have some code that uses the Oracle function add_months to increment a Date by X number of months.

I now need to re-implement the same logic in a C / C++ function. For reasons I don't want/need to go into I can't simply issue a query to oracle to get the new date.

Does anyone know of a simple and reliable way of adding X number of months to a time_t? Some examples of the types of calculations are shown below.

30/01/2009 + 1 month = 28/02/2009
31/01/2009 + 1 month = 28/02/2009
27/02/2009 + 1 month = 27/03/2009
28/02/2009 + 1 month = 31/03/2009
31/01/2009 + 50 months = 31/03/2013

Answer 1

Really new answer to a really old question!

Using this free and open source library, and a C++14 compiler (such as clang) I can now write this:

#include "date.h"

constexpr
date::year_month_day
add(date::year_month_day ymd, date::months m) noexcept
{
    using namespace date;
    auto was_last = ymd == ymd.year()/ymd.month()/last;
    ymd = ymd + m;
    if (!ymd.ok() || was_last)
        ymd = ymd.year()/ymd.month()/last;
    return ymd;
}

int
main()
{
    using namespace date;
    static_assert(add(30_d/01/2009, months{ 1}) == 28_d/02/2009, "");
    static_assert(add(31_d/01/2009, months{ 1}) == 28_d/02/2009, "");
    static_assert(add(27_d/02/2009, months{ 1}) == 27_d/03/2009, "");
    static_assert(add(28_d/02/2009, months{ 1}) == 31_d/03/2009, "");
    static_assert(add(31_d/01/2009, months{50}) == 31_d/03/2013, "");
}

And it compiles.

Note the remarkable similarity between the actual code, and the OP's pseudo-code:

30/01/2009 + 1 month = 28/02/2009
31/01/2009 + 1 month = 28/02/2009
27/02/2009 + 1 month = 27/03/2009
28/02/2009 + 1 month = 31/03/2009
31/01/2009 + 50 months = 31/03/2013

Also note that compile-time information in leads to compile-time information out.

C++20 <chrono> update:

Your C++ vendors are starting to ship C++20 <chrono> which can do this without a 3rd party library with nearly identical syntax.

Also note the somewhat unusual rules the OP requires for adding months, which is easily implementable in <chrono>:

If the resultant month overflows the day field or if the input month is the last day of the month, then snap the result to the end of the month.

#include <chrono>

constexpr
std::chrono::year_month_day
add(std::chrono::year_month_day ymd, std::chrono::months m) noexcept
{
    using namespace std::chrono;
    auto was_last = ymd == ymd.year()/ymd.month()/last;
    ymd = ymd + m;
    if (!ymd.ok() || was_last)
        ymd = ymd.year()/ymd.month()/last;
    return ymd;
}

int
main()
{
    using namespace std::chrono;
    static_assert(add(30d/01/2009, months{ 1}) == 28d/02/2009);
    static_assert(add(31d/01/2009, months{ 1}) == 28d/02/2009);
    static_assert(add(27d/02/2009, months{ 1}) == 27d/03/2009);
    static_assert(add(28d/02/2009, months{ 1}) == 31d/03/2009);
    static_assert(add(31d/01/2009, months{50}) == 31d/03/2013);
}

Demo.

Answer 2

You can use Boost.GregorianDate for this.

More specifically, determine the month by adding the correct date_duration, and then use end_of_month_day() from the date algorithms

Answer 3

Convert time_t to struct tm, add X to month, add months > 12 to years, convert back. tm.tm_mon is an int, adding 32000+ months shouldn't be a problem.

[edit] You might find that matching Oracle is tricky once you get to the harder cases, like adding 12 months to 29/02/2008. Both 01/03/2009 and 28/02/2008 are reasonable.

Answer 4

Method AddMonths_OracleStyle does what you need.

Perhaps you would want to replace IsLeapYear and GetDaysInMonth to some librarian methods.

#include <ctime>
#include <assert.h>

bool IsLeapYear(int year) 
{
    if (year % 4 != 0) return false;
    if (year % 400 == 0) return true;
    if (year % 100 == 0) return false;
    return true;
}

int daysInMonths[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int GetDaysInMonth(int year, int month)
{
    assert(month >= 0);
    assert(month < 12);

    int days = daysInMonths[month];

    if (month == 1 && IsLeapYear(year)) // February of a leap year
        days += 1;

    return days;
}

tm AddMonths_OracleStyle(const tm &d, int months)
{
    bool isLastDayInMonth = d.tm_mday == GetDaysInMonth(d.tm_year, d.tm_mon);

    int year = d.tm_year + months / 12;
    int month = d.tm_mon + months % 12;

    if (month > 11)
    {
        year += 1;
        month -= 12;
    }

    int day;

    if (isLastDayInMonth)
        day = GetDaysInMonth(year, month); // Last day of month maps to last day of result month
    else
        day = std::min(d.tm_mday, GetDaysInMonth(year, month));

    tm result = tm();

    result.tm_year = year;
    result.tm_mon = month;
    result.tm_mday = day;

    result.tm_hour = d.tm_hour;
    result.tm_min = d.tm_min;
    result.tm_sec = d.tm_sec;

    return result;
}

time_t AddMonths_OracleStyle(const time_t &date, int months)
{
    tm d = tm();

    localtime_s(&d, &date);

    tm result = AddMonths_OracleStyle(d, months);

    return mktime(&result);
}