Calculating the day of the week
|
The Gregorian Calendar runs on a 400-year cycle. Leap years are calculated with:
A leap year is a year wholly divided by 4
Unless it is wholly divided by 100, then it is not a leap year
Unless it is wholly divided by 400, then it is a leap year
So: 1600, 2000, 2400 are leap years, but 1900, 2100 etc are not.
Over 400 years there are 97 leap days, so the day of the week advances 497 times,
497/7 gives a remainder of 0, so over the 400-year Gregorian cycle the day of the
week resets to the same day. This means that the day of the week can be calculated
by reducing the year to a 0-399 offset within any 400-year cycle.
The standard modification of Zeller's formula to calculate the day of the week is:
offset[]={7,3,2,5,0,3,5,1,4,6,2,4}
if (month < 3) then year=year-1
day=(year + year/4 - year/100 + year/400 + offset[month] + day) MOD 7
|
|
| Incrementing a calendar date
|
This code will increment a date, correctly rolling over from the last
day of a month to the first day of the next month, or the next year.
| |
|
| Optimised *60 and *100
|
When manipulating clock times, you often need to multiply by 60 and 100, for
example centiseconds=((hours*60+minutes)*60+seconds)*100.
This is a small snippet of optimised code to do multiplication by 60 or 100.
There are 360000 = &057E40 centiseconds in a day, so this fits into a
3-byte 24-bit number.
| |
|
| Divide into hrs/min/sec/cs
|
This is simple unoptimised code to divide a centisecond count into hours,
minutes, seconds, and centiseconds; the reverse of Times60.
| |
| |
Authored by J.G.Harston