Convert unix timestamp to human readable date - c++

Question

Is there a contemporary way to convert unix timestamps to a human readable date?
Since I want to circumnavigate the year 2038 problem, I want to use int64s.
I target to convert e. g. 1205812558 to
year = 2008, month = 3, day = 18,
hour = 17, minute = 18, second = 36
All I have is now
auto year = totalSeconds / secondsPerYear + 1970;
// month and day missing
auto hours = totalSeconds / 3600 % 24;
auto minutes = totalSeconds / 60 % 60;
auto seconds = totalSeconds % 60;

Answer 1

In C++20 (according to the draft-spec for C++20 as it stands today), you will be able to say:
#include <chrono>
#include <iostream>
int
main()
{
using namespace std;
using namespace std::chrono;
cout << sys_seconds{1205812558s} << '\n';
cout << sys_seconds{32879409516s} << '\n';
}
and it will output:
2008-03-18 03:55:58
3011-11-28 17:18:36
These are datetimes in UTC.
You can use Howard Hinnant's date library to experiment with this extended <chrono> functionality today by adding:
#include "date/date.h"
and
using namespace date;
to the above program. You can experiment online with this program here.
A comment below asks for what this looks like if the value is stored in uint64_t. The answer is that you need to convert the integral type to seconds, and then the seconds to sys_seconds:
uint64_t i = 1205812558;
cout << sys_seconds{seconds(i)} << '\n';
There do exist limits on this contemporary functionality, but they live out near the years +/-32K (far beyond the limits of the accuracy of the current civil calendar).
To be completely transparent, there do exist ways of doing this using only C++98/11/14/17, but they are more complicated than this, and are subject to multithreading bugs. This is due to the use of an antiquated C API that was designed before things like multithreading and C++ were on the horizon, and when the year 2001 was only associated with science fiction (e.g. gmtime).

Answer 2

Wrapper
#include <chrono>
char* get_time(time_t unix_timestamp)
{
char time_buf[80];
struct tm ts;
ts = *localtime(&unix_timestamp);
strftime(time_buf, sizeof(time_buf), "%a %Y-%m-%d %H:%M:%S %Z", &ts);
return time_buf;
}

Answer 3

Howard Hinnant's date library makes things pretty easy:
#include "date.h"
int main()
{
using namespace date;
time_t time = 32879409516;
auto sysTime = std::chrono::system_clock::from_time_t(time);
auto date = year_month_day(floor<days>(sysTime));
std::cout << date << "\n";
}

Answer 4

A good straight forward solution but could do with some minor changes:
uint32_t days = (uint32_t)floor(subt / 86400);
uint32_t hours = (uint32_t)floor(((subt - days * 86400) / 3600) % 24);
uint32_t minutes = (uint32_t)floor((((subt - days * 86400) - hours * 3600) / 60) % 60);
uint32_t seconds = (uint32_t)floor(((((subt - days * 86400) - hours * 3600) - minutes * 60)) % 60);
printf("Time remaining: %u Days, %u Hours, %u Minutes, %u Seconds\n", days, hours, minutes, seconds);