How to convert the negative timestamp to a string. C++

Question

I have the negative timestamp (that is date older 1970, e.g. 15.04.1896). How to convert the given timestamp to the correct date string.

As I'm trying to do it

#include <ctime>
#include <string>
#include <iostream>


int _tmain(int argc, _TCHAR* argv[])
{
    time_t t = std::atol("-2326924800");
    struct tm * ptm;
    ptm = gmtime ( &t );

    std::cout << ptm->tm_year;
    std::cin.get();
    return 0;
}

Answer 1

Since it looks like you're using Windows (I'm inferring it from TCHAR), you probably want to use FileTimeToSystemTime(). It works with years since 1601.

Example:

#include <windows.h>
#include <stdio.h>
#include <string.h>

int main(void)
{
  SYSTEMTIME sysTime;
  FILETIME fileTime;
  long long seconds;

  sysTime.wYear = 1896;
  sysTime.wMonth = 4;
  sysTime.wDayOfWeek = 0;
  sysTime.wDay = 15;
  sysTime.wHour = 0;
  sysTime.wMinute = 0;
  sysTime.wSecond = 0;
  sysTime.wMilliseconds = 0;

  if (SystemTimeToFileTime(&sysTime, &fileTime))
  {
    seconds = *(long long*)&fileTime;
    seconds /= 10000000; // 100-nanoseconds to seconds since Jan 1st 1601
    seconds -= 11644473600; // 1601 to 1970
    printf("%d.%d.%d is %lld seconds from Jan 1st 1970\n",
           sysTime.wDay,
           sysTime.wMonth,
           sysTime.wYear,
           seconds);
  }
  else
  {
    printf("SystemTimeToFileTime() failed with error 0x%X\n", GetLastError());
  }

  // Now, convert it back...

  seconds += 11644473600; // 1970 to 1601
  seconds *= 10000000; // seconds since Jan 1st 1601 to 100-nanoseconds
  *(long long*)&fileTime = seconds;

  memset(&sysTime, 0, sizeof(sysTime));

  if (FileTimeToSystemTime(&fileTime, &sysTime))
  {
    seconds /= 10000000; // 100-nanoseconds to seconds since Jan 1st 1601
    seconds -= 11644473600; // 1601 to 1970
    printf("%lld seconds from Jan 1st 1970 is %d.%d.%d\n",
           seconds,
           sysTime.wDay,
           sysTime.wMonth,
           sysTime.wYear);
  }
  else
  {
    printf("FileTimeToSystemTime() failed with error 0x%X\n", GetLastError());
  }

  return 0;
}

Output:

15.4.1896 is -2326147200 seconds from Jan 1st 1970
-2326147200 seconds from Jan 1st 1970 is 15.4.1896

A rough estimation (1896-1970+3.5/12)*365.2425*24*3600 gives -2326010337. So, we're good.

EDIT:

If you want a DIY solution without involving any OS- or compiler-specific functions, use my seconds to date converter from this answer.

This snippet shows how to use it:

  struct tm t;
  SecondsSinceEpochToDateTime(&t, -2326924800LL);
  printf("-2326924800 is %d.%d.%d %d:%d:%d\n",
         t.tm_mday,
         t.tm_mon + 1,
         t.tm_year + 1900,
         t.tm_hour,
         t.tm_min,
         t.tm_sec);

And this is the output of it:

-2326924800 is 6.4.1896 0:0:0

Yeah, -2326924800 is not 15.4.1896 by the Gregorian calendar. It's 6.4.

Answer 2

I'd suggest using boost::posix_time.

std::cout << boost::posix_time::to_simple_string(boost::posix_time::from_time_t(0) + boost::posix_time::seconds(-2326924800));

time_t is unsuitable because it is not specified to accept negative values (see What is ultimately a time_t typedef to?)