查找给定的日期，月，月和月份中的“ n”次出现

Question

In order that a device (with limited memory) is able to manage its own timezone and daylight savings, I'm trying to calculate daylight savings triggers for 85 time zones based on a simplified description of each timezone. I have access to minimal C and C++ libraries within the device. The format of the timezone (inc. DST) description for each time zone is as follows:

• UTC - the base time and date from system clock
• GMTOffsetMinutes - offset from GMT with DST inactive
• DSTDeltaMinutes - modifier to above with DST active (as applicable to TZ)
• DSTStartMonth - month in which DST becomes active
• DSTStartNthOccurranceOfDay - the nth occurrence of the day name in month
• DSTDayOfWeek - Sun = 0 through to Sat = 6
• DSTStartHour - hour at which DST becomes active
• DSTStartMinute - minute at which DST becomes active
• and corresponding EndMonth, EndNth..., EndHour, EndMinute

I have found numerous examples going the other way, ie starting with the date, but they involve using the modulus, keeping the remainder and dropping the quotient hence I have been unable to transpose the formula to suit my needs.

I also tried to reuse the standard "Jan = 6, Feb = 2, Mar = 2, Apr = 5, May = 0, etc. modifier table and year modifiers from the "tell me what day the 25th of June, 2067 is?" party trick and developed the following algorithm.

Date = DayOfWeek + ((NthOccuranceOfDay - 1) x 7 ) - MonthCode - YearCode

This worked for the first 6 random test dates I selected but then I started to see dates for which it failed. Is it possible that the basic algorithm is sound but I'm missing a further modifier or maybe that I'm applying the modifiers incorrectly?

Is there another solution I could utilize?

Answer 1

Using this open source, cross platform date library , one can write:

#include "date.h"
#include <iostream>

int
main()
{
    using namespace date;
    year_month_day us_daylight_starts = sys_days(sun[2]/mar/2015);
    year_month_day us_daylight_ends   = sys_days(sun[1]/nov/2015);
    std::cout << us_daylight_starts << '\n';
    std::cout << us_daylight_ends << '\n';
}

which will output:

2015-03-08
2015-11-01

The formulas this library is based on are in the public domain and documented here .

The algorithms paper has very complete unit tests validating the date algorithms over a range of millions of years (a far larger range than is necessary).

Sometimes daylight savings rules are written in terms of the last weekday of a month. That is just as easily handled:

year_month_day ymd = sys_days(sun[last]/nov/2015);
std::cout << ymd << '\n';  // 2015-11-29

Answer 2

That formula will be off by one week (or even two weeks) if MonthCode + YearCode is greater than or equal to DayOfWeek , because in that case you will be counting NthOccurenceOfDay from a negative date.

As an alternative, with no tables, you can compute the day of week of the first of the month using, for example, Zeller's algorithm :

int NthOccurrence(int year, int month, int n, int dayOfWeek) {
  // year is the current year (eg. 2015)
  // month is the target month (January == 1...December == 12)
  // Finds the date of the nth dayOfWeek (Sun == 0...Sat == 6)

  // Adjust month and year
  if (month < 3) { --year, month += 12; }
  // The gregorian calendar is a 400-year cycle
  year = year % 400;
  // There are no leap years in years 100, 200 and 300 of the cycle.
  int century = year / 100;
  int leaps = year / 4 - century;
  // A normal year is 52 weeks and 1 day, so the calendar advances one day.
  // In a leap year, it advances two days.
  int advances = year + leaps;
  // This is either magic or carefully contrived,
  // depending on how you look at it:
  int month_offset = (13 * (month + 1)) / 5;
  // From which, we can compute the day of week of the first of the month:
  int first = (month_offset + advances) % 7;
  // If the dayOfWeek we're looking for is at least the day we just
  // computed, we just add the difference. Otherwise, we need to add 7.
  // Then we just add the desired number of weeks.
  int offset = dayOfWeek - first;
  if (offset < 0) offset += 7;
  return 1 + offset + (n - 1) * 7;
}