C++ get start of this current year in unixtime - c++

Question

I'm looking for a way to retrieve the start of the current year as a unix timestamp.
For example if we're on 2017-10-16 the unix timestamp is 1523318400. I have to retrieve 1483228800 (2017-01-01) instead. And it must work for the next years too of course.

Answer 1

There are functions to add and subtract a number of months, days, minutes and seconds from a time_t, which can be used to calculate the the time_t for a point in the past, but it looks quite awkward to find the correct number of units to remove. (cpp reference : time_point). I also looked at original C function mktime. However, whilst creating a time_t and then a struct tm*, the issue is correctly generating a timezone correct version.
So my solution is something like this....
int getYear(time_t now)
{
struct tm * tnow = std::gmtime(&now);
return tnow->tm_year + 1900;
}
std::time_t calculateYear( int currentYear )
{
int epochYear = currentYear - 1970;
int leapYears = (epochYear + 1) / 4;
time_t result = epochYear * 24 * 60 * 60 * 365;
result += leapYears * 24 * 60 * 60;
return result;
}
The code is good for years between 1970 (first time_t value) and 2100, which is not a leap year from a 100 year rule.
The number of leap years is strange, as whilst 2012 is a leap year, it is 2013 which is the first year beginning to count it.

Answer 2

You could use Howard Hinnant's free, open-source C++11/14/17 date/time library. It would be as simple as this:
#include "date/date.h"
date::sys_seconds
start_of_year(date::sys_seconds t)
{
using namespace date;
return sys_days{year_month_day{floor<days>(t)}.year()/jan/1};
}
You could use it like this:
#include <iostream>
int
main()
{
using date::operator<<;
using namespace std::chrono_literals;
std::cout << start_of_year(date::sys_seconds{1523318400s}) << '\n';
}
This outputs:
1514764800s
Note that this is not the answer you said you wanted. However it is correct. You can debug this discrepancy with this library as well:
std::cout << date::sys_seconds{1523318400s} << '\n';
This outputs:
2018-04-10 00:00:00
instead of 2017-10-16. You can find the Unix Time stamp for 2017-10-16 with:
using namespace date::literals;
std::cout << date::sys_seconds{date::sys_days{2017_y/10/16}}.time_since_epoch() << '\n';
which outputs:
1508112000s
And:
std::cout << start_of_year(date::sys_seconds{1508112000s}).time_since_epoch() << '\n';
will output:
1483228800s
You can also use this library to find the current year:
date::year
current_year()
{
using namespace date;
using namespace std::chrono;
return year_month_day{floor<days>(system_clock::now())}.year();
}
And you could rewrite (or overload) start_of_year to take a date::year instead of (or in addition to) date::sys_seconds:
date::sys_seconds
start_of_year(date::year y)
{
using namespace date;
return sys_days{y/jan/1};
}
And now you can write:
int
main()
{
using date::operator<<;
std::cout << start_of_year(current_year()).time_since_epoch() << '\n';
}
which currently outputs:
1483228800s