8d67a1ce88
Britain and its colonies eliminated 11 days (not 10), following September 2, 1752. From "A.D. 1751. Anno vicesimo quarto GEORGII II. CAP. XXIII. An Act for Regulating the Commencement of the Year; and for Correcting the Calendar now in Use.": "... and that the natural Day next immediately following the said 2nd Day of *September* [1752], shall be called, reckoned and accounted to be the 14th Day of *September*, omitting for that Time only the 11 intermediate nominal Days of the common Calendar; and that the several natural Days, which shall follow and succeed next after the said 14th Day of *September*, shall be respectively called, reckoned and numbered forwards in numerical Order from the said 14th Day of *September*, according to the Order and Succession of Days now used in the present Calendar; " Added a caution note on using cal for very old dates. Problem mentionned in PR 5215 by John Franklin (franklin@bev.net). Thanks to Perry Metzger for his comments and for reviewing this man page. |
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cal.1 | ||
cal.c | ||
Makefile | ||
README |
The cal(1) date routines were written from scratch, basically from first principles. The algorithm for calculating the day of week from any Gregorian date was "reverse engineered". This was necessary as most of the documented algorithms have to do with date calculations for other calendars (e.g. julian) and are only accurate when converted to gregorian within a narrow range of dates. 1 Jan 1 is a Saturday because that's what cal says and I couldn't change that even if I was dumb enough to try. From this we can easily calculate the day of week for any date. The algorithm for a zero based day of week: calculate the number of days in all prior years (year-1)*365 add the number of leap years (days?) since year 1 (not including this year as that is covered later) add the day number within the year this compensates for the non-inclusive leap year calculation if the day in question occurs before the gregorian reformation (3 sep 1752 for our purposes), then simply return (value so far - 1 + SATURDAY's value of 6) modulo 7. if the day in question occurs during the reformation (3 sep 1752 to 13 sep 1752 inclusive) return THURSDAY. This is my idea of what happened then. It does not matter much as this program never tries to find day of week for any day that is not the first of a month. otherwise, after the reformation, use the same formula as the days before with the additional step of subtracting the number of days (11) that were adjusted out of the calendar just before taking the modulo. It must be noted that the number of leap years calculation is sensitive to the date for which the leap year is being calculated. A year that occurs before the reformation is determined to be a leap year if its modulo of 4 equals zero. But after the reformation, a year is only a leap year if its modulo of 4 equals zero and its modulo of 100 does not. Of course, there is an exception for these century years. If the modulo of 400 equals zero, then the year is a leap year anyway. This is, in fact, what the gregorian reformation was all about (a bit of error in the old algorithm that caused the calendar to be inaccurate.) Once we have the day in year for the first of the month in question, the rest is trivial.