C++ Add Months to Chrono::System_Clock::Time_Point

C++ Add months to chrono::system_clock::time_point

Overview

This is a very interesting question with several answers. The "correct" answer is something you must decide for your specific application.

With months, you can choose to do either chronological computations or calendrical computations. A chronological computation deals with regular units of time points and time durations, such as hours, minutes and seconds. A calendrical computation deals with irregular calendars that mainly serve to give days memorable names.

The Chronological Computation

If the question is about some physical process months in the future, physics doesn't care that different months have different lengths, and so a chronological computation is sufficient:

• The baby is due in 9 months.

• What will the weather be like here 6 months from now?

In order to model these things, it may be sufficient to work in terms of the average month. One can create a std::chrono::duration that has precisely the length of an average Gregorian (civil) month. It is easiest to do this by defining a series of durations starting with days:

days is 24 hours:

using days = std::chrono::duration
    <int, std::ratio_multiply<std::ratio<24>, std::chrono::hours::period>>;

years is 365.2425 days, or ¹⁴⁶⁰⁹⁷/₄₀₀ days:

using years = std::chrono::duration
    <int, std::ratio_multiply<std::ratio<146097, 400>, days::period>>;

And finally months is ¹/₁₂ of years:

using months = std::chrono::duration
    <int, std::ratio_divide<years::period, std::ratio<12>>>;

Now you can easily compute 8 months from now:

auto t = system_clock::now() + months{8};

Important note: This computation does not preserve the time of day, or even the day of the month.

The Calendrical Computation

It is also possible to add months while preserving time of day and day of month. Such computations are calendrical computations as opposed to chronological computations.

After choosing a calendar (such as the Gregorian (civil) calendar, the Julian calendar, or perhaps the Islamic, Coptic or Ethiopic calendars — they all have months, but they are not all the same months), the process is:

1. Convert the system_clock::time_point to the calendar.

2. Perform the months computation in the calendrical system.

3. Convert the new calendar time back into system_clock::time_point.

You can use Howard Hinnant's free, open-source date/time library to do this for a few calendars. Here is what it looks like for the civil calendar:

#include "date.h"

int
main()
{
    using namespace date;
    using namespace std::chrono;

    // Get the current time
    auto now = system_clock::now();
    // Get a days-precision chrono::time_point
    auto sd = floor<days>(now);
    // Record the time of day
    auto time_of_day = now - sd;
    // Convert to a y/m/d calendar data structure
    year_month_day ymd = sd;
    // Add the months
    ymd += months{8};
    // Add some policy for overflowing the day-of-month if desired
    if (!ymd.ok())
        ymd = ymd.year()/ymd.month()/last;
    // Convert back to system_clock::time_point
    system_clock::time_point later = sys_days{ymd} + time_of_day;
}

For grins I just ran this, and compared it with now + months{8} and got:

now   is           2017-03-25 15:17:14.467080
later is           2017-11-25 15:17:14.467080  // calendrical computation
now + months{8} is 2017-11-24 03:10:02.467080  // chronological computation

This gives a rough "feel" for how the calendrical computation differs from the chronological computation. The latter is perfectly accurate on average; it just has a deviation from the calendrical on the order of a few days. And sometimes the simpler (latter) solution is close enough, and sometimes it is not. Only you can answer that question.

The Calendrical Computation — Now with timezones

Finally, you might want to perform your calendrical computation in a specific timezone. The previous computation was UTC.

Side note: system_clock is not specified to be UTC, but the de facto standard is that it is Unix Time which is a very close approximation to UTC.

You can use Howard Hinnant's free, open-source timezone library to do this computation. This is an extension of the previously mentioned datetime library.

The code is very similar, you just need to convert to local time from UTC, then to a local calendar, do the computation then back to local time, and finally back to system_clock::time_point (UTC):

#include "tz.h"

int
main()
{
    using namespace date;
    using namespace std::chrono;

    // Get the current local time
    auto lt = make_zoned(current_zone(), system_clock::now());
    // Get a days-precision chrono::time_point
    auto ld = floor<days>(lt.get_local_time());
    // Record the local time of day
    auto time_of_day = lt.get_local_time() - ld;
    // Convert to a y/m/d calendar data structure
    year_month_day ymd{ld};
    // Add the months
    ymd += months{8};
    // Add some policy for overflowing the day-of-month if desired
    if (!ymd.ok())
        ymd = ymd.year()/ymd.month()/last;
    // Convert back to local time
    lt = local_days{ymd} + time_of_day;
    // Convert back to system_clock::time_point
    auto later = lt.get_sys_time();
}

Updating our results I get:

now   is           2017-03-25 15:17:14.467080
later is           2017-11-25 15:17:14.467080  // calendrical: UTC
later is           2017-11-25 16:17:14.467080  // calendrical: America/New_York
now + months{8} is 2017-11-24 03:10:02.467080  // chronological computation

The time is an hour later (UTC) because I preserved the local time (11:17am) but the computation started in daylight saving time, and ended in standard time, and so the UTC equivalent is later by 1 hour.

I used current_zone() to pick up my current location, but I could have also used a specific time zone (e.g. "Asia/Tokyo").

C++20 Update

As I write this update, technical work has ceased on C++20, and it looks like we will have a new C++ standard later this year (just administrative work left to do to complete C++20).

The advice in this answer translates well to C++20:

1. For the chronological computation, std::chrono::months is supplied by <chrono> so you don't have to compute it yourself.

2. For the UTC calendrical computation, loose #include "date.h" and use instead #include <chrono>, and drop using namespace date, and things will just work.

3. For the time zone sensitive calendrical computation, loose #include "tz.h" and use instead #include <chrono>, drop using namespace date, and replace make_zoned with zoned_time, and you're good to go.

How do I add a number of days to a date in C++20 chrono?

A "date" in <chrono> is just a time point with days-precision. And a
"date-time" is just a chrono::time_point (usually based on
system_clock) with precision finer than days. And the canonical "date type" in
<chrono> is:

chrono::time_point<chrono::system_clock, chrono::days>

This is nothing but a days-precision time_point based on
system_clock. This particular time_point also has a convenience
type alias: sys_days.

using sys_days = time_point<system_clock, days>;

So to add a number of days to sys_days one just does:

sys_days tp = ...;
tp += days{n};
// or
auto tp2 = tp + days{n};

If it is just one day, you can also just:

++tp;

This is very efficient because under the hood it is just adding two
ints.

If your time_point has precision finer than days, it is still the
same procedure to add days to it:

auto tp = system_clock::now() + days{n};

When I try this I get a compile-time error:

auto d = July/4/2020;
auto d2 = d + days{5};
          ~ ^ ~~~~~~~
error: invalid operands to binary expression
   ('std::chrono::year_month_day' and 'std::chrono::days')

The variable d above has type year_month_day. Even though year_month_day is semantically equivalent to sys_days,
it does not directly support days-precision arithmetic.
year_month_day is a days-precision time point (but not a
chrono::time_point). Instead it is a {year, month, day} data
structure. You can easily perform days-precision arithmetic by
converting it to sys_days first:

auto d = July/4/2020;
auto d2 = sys_days{d} + days{5};

In this case, the type of the result d2 is sys_days. If you would like the
result to have year_month_day, then just convert it back:

year_month_day d2 = sys_days{d} + days{5};

Rationale for this design:

Efficiency. The conversion from year_month_day to sys_days (and
back) takes a bit of number crunching. It isn't a huge amount. But
it is large compared to a single integral addition.

If the <chrono> library provided days-precision arithmetic for year_month_day,
the algorithm would be to convert the year_month_day to sys_days,
do the arithmetic, and then convert the result back to
year_month_day. If you have a lot of days-precision arithmetic to
do, it is better to just traffic in sys_days, and convert to
year_month_day only when necessary (i.e. when you need to get the
year, month or day fields out of it).

If the library provided the day-precision arithmetic for year_month_day,
clients would blindly use it, not realizing that year_month_day is the
wrong data structure for their application. It would be akin to giving
std::list an indexing operator (which would be quite easy to do):

template <class T, class A>
T&
list<T, A>::operator[](size_type i)
{
    return *advance(begin(), i);
}

Providing such an API just makes it too easy to write inefficient
code. sys_days is the data structure of choice for doing days
precision arithmetic.

It still doesn't work for me:

auto d = July/4/2020;
auto d2 = sys_days{d} + day{5};
          ~~~~~~~~~~~ ^ ~~~~~~
error: invalid operands to binary expression
   ('std::chrono::sys_days' and 'std::chrono::day')

You need to use days, not day. day is a calendrical specifier
for the day of a month in the civil calendar. days is a
chrono::duration. See this stack overflow Q/A for a more in-depth
discussion about the distinction between these two concepts.

If I have a year_month_day and I want to add a few days to it that
I know won't reach the end of the month, can I do that without
converting to sys_days and thus gain efficiency?

Yes.

auto d = July/4/2020;
auto d2 = d.year()/d.month()/(d.day() + days{5});

The above simply adds 5 to the day field of d. There is no
checking to see if the result goes past the last day of the month. If
it does go past the last day of the month, that result will be stored
in the day field (up to day 255), and d2.ok() will return false.

---

year_month_day is good for retrieving the year, month or day
fields from a date. It is also good for years and months
precision calendrical arithmetic. sys_days is good for days
precision arithmetic, and for converting to a finer precision
(a date-time).

year_month_day and sys_days can convert to one another with no
information loss. Use whichever data structure makes the most sense.
And if you forget, the compiler will remind you, just like it does for
vector (no push_front), list (no indexing operator), and
forward_list (no size).

Add time duration to C++ timepoint

If you want to add five hours to startTimePoint, it's boringly simple:

startTimePoint += hours(5); // from the alias std::chrono::hours

Live example.

By the way, you're trying to convert a steady_clock::now() into a system_clock::time_point, which shouldn't even compile. Change the steady_clock::now() to system_clock::now() and you should be good to go.

How to add chrono::year_month_day to chrono::sys_seconds

This is chrono catching logic bugs for you at compile-time. Adding date2 + calDuration is akin to adding tomorrow + today. It just doesn't make sense. And that's why it is a compile-time error.

What you may mean is that you have durations years, months and days. This is not the same as the similarly named types year, month and day. The plural forms are chrono::durations, just like minutes and nanoseconds. days is 24h. And years and months are the average length of those units in the civil calendar.

Conversely, the singular forms year, month and day are the calendrical components that give a name to a day in the civil calendar, e.g. 2020y, December, and 18d. And it is these singular forms that make up a year_month_day, e.g. 2020y/December/18d.

See this SO answer for a deep-dive on the difference between month and months.

There are multiple ways to add the units years and months to a time_point. See this SO answer for a deep-dive on that topic.

Creating a `std::chrono::time_point` from a calendar date known at compile time

Yes, you can do the entire computation at compile time, creating a constexpr system_clock::time_point using Howard Hinnant's date/time library.

#include "date/date.h"
#include <chrono>

int
main()
{
    using namespace date;
    using namespace std::chrono;
    constexpr system_clock::time_point tp = sys_days{January/9/2014} + 12h + 35min + 34s;
    static_assert(tp == system_clock::time_point{1389270934s}, "");
}

This is assuming that the date/time is UTC. If it isn't, you will have to manually add/subtract the UTC offset to make it so. As time zone rules are changed at the whim of politicians all the time, there is little hope in making them constexpr. Even historical time zone rules are updated when misunderstandings come to light.

Also this program will port to C++20 by dropping #include "date/date.h" and using namespace date;. Also using Howard Hinnant's date/time library requires C++14 constexpr muscle. C++11 constexpr is not sufficient (but you can do it at run-time, dropping the constexpr and static_assert).

Extract year/month/day etc. from std::chrono::time_point in C++

You can only extract this information from a system_clock::time_point. This is the only system-supplied clock that has a relationship with the civil calendar. Here is how to get the current time_point using this clock:

 system_clock::time_point now = system_clock::now();

You can then convert this to a time_t with:

time_t tt = system_clock::to_time_t(now);

Using the C library you can then convert a time_t to a tm, but you must choose whether you want that conversion to happen in the UTC timezone, or you local timezone:

tm utc_tm = *gmtime(&tt);
tm local_tm = *localtime(&tt);

Then you can print out the components of the tm, for example:

std::cout << local_tm.tm_year + 1900 << '\n';
std::cout << local_tm.tm_mon + 1 << '\n';
std::cout << local_tm.tm_mday << '\n';

Additionally

If you want, you can take advantage of this non-guaranteed information:

Every implementation of system_clock I'm aware of is based on unix time. I.e. the number of seconds since New Years 1970 UTC, neglecting leap seconds. And the precision of this count is usually finer than seconds. Here is a complete program which extracts all of this information:

#include <chrono>
#include <ctime>
#include <iostream>

int
main()
{
    using namespace std;
    using namespace std::chrono;
    typedef duration<int, ratio_multiply<hours::period, ratio<24> >::type> days;
    system_clock::time_point now = system_clock::now();
    system_clock::duration tp = now.time_since_epoch();
    days d = duration_cast<days>(tp);
    tp -= d;
    hours h = duration_cast<hours>(tp);
    tp -= h;
    minutes m = duration_cast<minutes>(tp);
    tp -= m;
    seconds s = duration_cast<seconds>(tp);
    tp -= s;
    std::cout << d.count() << "d " << h.count() << ':'
              << m.count() << ':' << s.count();
    std::cout << " " << tp.count() << "["
              << system_clock::duration::period::num << '/'
              << system_clock::duration::period::den << "]\n";

    time_t tt = system_clock::to_time_t(now);
    tm utc_tm = *gmtime(&tt);
    tm local_tm = *localtime(&tt);
    std::cout << utc_tm.tm_year + 1900 << '-';
    std::cout << utc_tm.tm_mon + 1 << '-';
    std::cout << utc_tm.tm_mday << ' ';
    std::cout << utc_tm.tm_hour << ':';
    std::cout << utc_tm.tm_min << ':';
    std::cout << utc_tm.tm_sec << '\n';
}

It is handy to create a custom duration to model days:

typedef duration<int, ratio_multiply<hours::period, ratio<24> >::type> days;

Now you can get the time since the epoch, to as fine a precision as it can manage, with:

system_clock::duration tp = now.time_since_epoch();

Then truncate it to days, and subtract that off.

Then truncate it to hours, and subtract that off.

Continue until you've subtracted off the seconds.

What you're left with is the fraction of a second with the units of system_clock::duration. So print out that run time value and the compile time units of that value as shown.

For me this program prints out:

15806d 20:31:14 598155[1/1000000]
2013-4-11 20:31:14

My output indicates the system_clock::duration precision is microseconds. If desired, that can be truncated to milliseconds with:

milliseconds ms = duration_cast<milliseconds>(tp);

Update

This header-only C++11/14 library encapsulates the work above, reducing client work down to:

#include "date.h"
#include <iostream>

int
main()
{
    // Reduce verbosity but let you know what is in what namespace
    namespace C = std::chrono;
    namespace D = date;
    namespace S = std;

    auto tp = C::system_clock::now(); // tp is a C::system_clock::time_point
    {
        // Need to reach into namespace date for this streaming operator
        using namespace date;
        S::cout << tp << '\n';
    }
    auto dp = D::floor<D::days>(tp);  // dp is a sys_days, which is a
                                      // type alias for a C::time_point
    auto ymd = D::year_month_day{dp};
    auto time = D::make_time(C::duration_cast<C::milliseconds>(tp-dp));
    S::cout << "year        = " << ymd.year() << '\n';
    S::cout << "month       = " << ymd.month() << '\n';
    S::cout << "day         = " << ymd.day() << '\n';
    S::cout << "hour        = " << time.hours().count() << "h\n";
    S::cout << "minute      = " << time.minutes().count() << "min\n";
    S::cout << "second      = " << time.seconds().count() << "s\n";
    S::cout << "millisecond = " << time.subseconds().count() << "ms\n";
}

Which just output for me:

2015-07-10 20:10:36.023017
year        = 2015
month       = Jul
day         = 10
hour        = 20h
minute      = 10min
second      = 36s
millisecond = 23ms

Another Update

This library grew into a C++ standards proposal and is now in the C++20 working draft. The syntax for extracting these fields from a system_clock::time_point in C++20 will be:

#include <chrono>

int
main()
{
    using namespace std::chrono;
    auto tp = system_clock::now();
    auto dp = floor<days>(tp);
    year_month_day ymd{dp};
    hh_mm_ss time{floor<milliseconds>(tp-dp)};
    auto y = ymd.year();
    auto m = ymd.month();
    auto d = ymd.day();
    auto h = time.hours();
    auto M = time.minutes();
    auto s = time.seconds();
    auto ms = time.subseconds();
}

The above assumes you want these fields in UTC. If you prefer them in some other time zone, that will also be possible. For example, here is how to do it in your computer's current local time zone:

#include <chrono>

int
main()
{
    using namespace std::chrono;
    auto tp = zoned_time{current_zone(), system_clock::now()}.get_local_time();
    auto dp = floor<days>(tp);
    year_month_day ymd{dp};
    hh_mm_ss time{floor<milliseconds>(tp-dp)};
    auto y = ymd.year();
    auto m = ymd.month();
    auto d = ymd.day();
    auto h = time.hours();
    auto M = time.minutes();
    auto s = time.seconds();
    auto ms = time.subseconds();
}

The only difference above is the construction of tp which now has type local_time as opposed to sys_time in the UTC example. Alternatively one could have picked an arbitrary time zone with this small change:

auto tp = zoned_time{"Europe/London", system_clock::now()}.get_local_time();

std::chrono: add custom duration to time_point

The conversion fails, because there is no conversion from

std::chrono::time_point< std::chrono::system_clock,
                         std::chrono::system_clock::duration >

to

std::chrono::time_point< std::chrono::system_clock,
                         std::chrono::duration< double > >

The easiest way would be to give double_prec_seconds as a template parameter to time_point, see std::chrono::time_point

typedef std::chrono::time_point< std::chrono::system_clock,
                                 double_prec_seconds > timepoint_t;

then you already have the proper type for t3 and do_something.

Add time of day information to year_month_day with Howard Hinnant's date library

using namespace std::chrono;
auto tp = date::sys_days{X} + hours{A} + minutes{B} + seconds{C};

The type of tp is std::chrono::time_point<system_clock, seconds>, and represents a time point in UTC.

The only thing Howard Hinnant's date library adds in this example is the conversion from date::year_month_day to sys_days, which itself is just a typedef for time_point<system_clock, days>. After that conversion, you're working entirely within the C++11/14 <chrono> library.

Trying to subtract a std::chrono::duration from a time_point (C++), but my code isn't compiling. Any suggestions?

float numSeconds = 50;

The following line:

std::chrono::steady_clock::time_point startTime = std::chrono::steady_clock::now();

could be more concisely written as:

auto startTime = std::chrono::steady_clock::now();

Both syntaxes are equivalent and correct.

The reason this doesn't compile:

auto duration = std::chrono::duration<float, std::chrono::seconds>(numSeconds);

is that the second template argument to duration takes a std::ratio, not a std::chrono::duration. The correct way to say this is:

auto duration = std::chrono::duration<float, std::ratio<1, 1>>(numSeconds);

The above means that duration has type std::chrono::duration with a representation of float and a period of ¹/₁ seconds. That means that this is just a floating point count of seconds.

The second template parameter of std::ratio defaults to 1, so the above can be simplified to:

auto duration = std::chrono::duration<float, std::ratio<1>>(numSeconds);

And the second template parameter of std::chrono::duration defaults to std::ratio<1> so the above can be further simplified to:

auto duration = std::chrono::duration<float>(numSeconds);

The following doesn't compile:

startTime -= duration;

because chrono has a rule that says values with floating-point representation never implicitly convert to those with integral representation. This is to avoid the truncating error that results when one assigns a float to an int.

There are several ways to fix this. For example you could store the result in a float-based time_point:

auto anotherTime = startTime - duration;

anotherTime has type std::chrono::time_point<std::chrono::steady_clock, std::chrono::duration<float, std::chrono::steady_clock::period>>.

Or you could explicitly cast duration back to integral type:

startTime -= std::chrono::duration_cast<std::chrono::seconds>(duration);

If you choose the latter, then there really is no point in using float at all. And the entire sequence could look like:

using namespace std::chrono_literals;
auto startTime = std::chrono::steady_clock::now();
auto duration = 50s;
startTime -= duration;

or:

using namespace std::chrono_literals;
auto startTime = std::chrono::steady_clock::now();
startTime -= 50s;

or:

using namespace std::chrono_literals;
auto startTime = std::chrono::steady_clock::now() - 50s;

Here is a 1h video tutorial on <chrono> that you may find helpful: https://www.youtube.com/watch?v=P32hvk8b13M

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