Convert date/time to time-stamp and vice versa

问题:

I am trying to implement in C two simple convertors, date/time to time-stamp and vice versa, without any dependencies on time library routines, such as mktime, etc.

The time-stamp is in seconds, and the date/time structure is in the following format:

• unsigned char year: 0 to 99 (representing the range 2000 to 2099)

• unsigned char month: 1 to 12

• unsigned char day: 1 to 31

• unsigned char hour: 0 to 23

• unsigned char minute: 0 to 59

• unsigned char second: 0 to 59

I would like to have a second opinion on the dt2ts convertor (assuming that the input is legal):

unsigned int dt2ts(const dt_t* dt)
{
    static unsigned short days[] = {0,31,59,90,120,151,181,212,243,273,304,334};
    return ((((dt->year*365+dt->year/4)+days[dt->month-1]+dt->day)*24+dt->hour)*60+dt->minute)*60+dt->second;
}

In addition to that, I would appreciate some help completing the ts2dt convertor:

void ts2dt(unsigned int ts,dt_t* dt)
{
    dt->second = ts%60; ts /= 60;
    dt->minute = ts%60; ts /= 60;
    dt->hour   = ts%24; ts /= 24;
    dt->day    = ?????;
    dt->month  = ?????;
    dt->year   = ?????;
}

Thanks

回答1:

OP is all ready well handling the hours, minutes, seconds. Just a bit of assist on Y,M,D.

Note: The number of days from Jan 1, 2000 to Dec 31, 2099 needs at least a 16 bit integer. Following should work even if unsigned is 2 bytes.

unsigned DivRem(unsigned Dividend, unsigned Divisor, unsigned *Remainder) {
  unsigned Quotient = Dividend/Divisor;
  *Remainder = Dividend - Quotient*Divisor;
  return Quotient;
}

void Day2000ToYMD(unsigned DaySince2000Jan1, unsigned *Y, unsigned *M, unsigned *D) {
  unsigned OlympiadDay;  // Every 4 years is an Olympiad
  *Y = 4*DivRem(DaySince2000Jan1, 365*4+1, &OlympiadDay);
  *D = 1;
  if (OlympiadDay >= (31+29-1)) {  // deal with Feb 29th and after
    OlympiadDay--;
    if (OlympiadDay == (31+29-1)) {
      (*D)++;
    }
  }
  unsigned YearDay;      // Day of the year 0 to 364
  *Y += DivRem(OlympiadDay, 365, &YearDay);
  static const unsigned short days[] = {0,31,59,90,120,151,181,212,243,273,304,334,365};
  *M = 1;
  while (days[*M] <= YearDay) (*M)++;
  *D += YearDay - days[*M - 1];
}

---

[Edit] The answer provided tries to keep the concept of the year as Jan 1 to Dec 31. As this answer does not need to handle the leap year years about 100 and 400 years, I've kept with this style.

In general, once those 2 rules are added, the math becomes easier to if one shifts the beginning of the year to March 1 and ending on Feb 28/29. FWIW, this is a more consistent view of the ancient development of the Julian/Gregorian calendar. Thus *Oct*ober is then the 8th month and *Dec*ember is the 10th month.