Extracting Date and Time Components from std::chrono::time_point

Problem:

How can we retrieve specific date and time components, such as year, month, day, hours, minutes, seconds, and milliseconds, from an std::chrono::time_point object?

Solution:

To extract this information, we must utilize the system_clock::time_point as it's the only system-supplied clock that has a connection to the civil calendar.

<code class="cpp">system_clock::time_point now = system_clock::now();</code>

Converting to a Time Stamp:

Next, convert the time_point to a time_t:

<code class="cpp">time_t tt = system_clock::to_time_t(now);</code>

Extracting Components:

Using the C library, convert the time_t to a tm, specifying whether you want conversion in UTC or the local time zone:

<code class="cpp">tm utc_tm = *gmtime(&tt);
tm local_tm = *localtime(&tt);</code>

Now, you can print the components of the tm:

<code class="cpp">std::cout << local_tm.tm_year + 1900 << '\n';
std::cout << local_tm.tm_mon + 1 << '\n';
std::cout << local_tm.tm_mday << '\n';

Extended Extraction:

While system_clock is typically based on Unix time, its precision is often greater than seconds. To access the high-resolution components, use the following code:

<code class="cpp">using namespace std;
using namespace std::chrono;

int main()
{
    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;

    cout << d.count() << "d " << h.count() << ':'
         << m.count() << ':' << s.count();
    cout << " " << tp.count() << "["
         << system_clock::duration::period::num << '/'
         << system_clock::duration::period::den << "]\n";
}

By converting the time_point to days, hours, minutes, seconds, and extracting the fraction of a second, this code outputs:

15806d 20:31:14 598155[1/1000000]

Library Solutions:

The header-only date library simplifies this process, allowing the extraction with the following code:

<code class="cpp">#include "date.h"

int main()
{
    auto tp = std::chrono::system_clock::now();
    auto dp = date::floor<date::days>(tp);
    auto ymd = date::year_month_day{dp};
    auto time = date::make_time(std::chrono::duration_cast<std::chrono::milliseconds>(tp - dp));

    std::cout << "year        = " << ymd.year() << '\n';
    std::cout << "month       = " << ymd.month() << '\n';
    std::cout << "day         = " << ymd.day() << '\n';
    std::cout << "hour        = " << time.hours().count() << "h\n";
    std::cout << "minute      = " << time.minutes().count() << "min\n";
    std::cout << "second      = " << time.seconds().count() << "s\n";
    std::cout << "millisecond = " << time.subseconds().count() << "ms\n";
}

C 20 Integration:

In C 20, a proposed syntax is included for extracting these fields using system_clock::time_point:

<code class="cpp">#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();
}</code>

Additionally, you can specify a preferred time zone for the extraction.

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