22   int64_t years = *remainingDays / daysPerYears;  in computeRemainingYears()  local
23   if (years == quotientYears)  in computeRemainingYears()
24     years--;  in computeRemainingYears()
25   *remainingDays -= years * daysPerYears;  in computeRemainingYears()
26   return years;  in computeRemainingYears()
35 // makes it easier to count how many leap years have passed using division.
37 // While calculating numbers of years in the days, the following algorithm
38 // subdivides the days into the number of 400 years, the number of 100 years and
39 // the number of 4 years. These numbers of cycle years are used in calculating
41 // and isLeapYear(). Then compute the total number of years in days from these
44 // Compute the number of months from the remaining days. Finally, adjust years
88   // The remaining number of years after computing the number of  in update_from_seconds()
90   // years.  in update_from_seconds()
94   // The remaining number of years after computing the number of  in update_from_seconds()
95   // "hundred year cycles" will be 25 four year cycles or less in 100 years.  in update_from_seconds()
99   // The remaining number of years after computing the number of  in update_from_seconds()
100   // "four year cycles" will be 4 one year cycles or less in 4 years.  in update_from_seconds()
104   // Calculate number of years from year 2000.  in update_from_seconds()
105   int64_t years = remainingYears + 4 * numOfFourYearCycles +  in update_from_seconds()  local
126     years++;  in update_from_seconds()
129   if (years > INT_MAX || years < INT_MIN)  in update_from_seconds()
132   // All the data (years, month and remaining days) was calculated from  in update_from_seconds()
134   tm->tm_year = static_cast<int>(years + 2000 - TimeConstants::TIME_YEAR_BASE);  in update_from_seconds()