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This commit is contained in:
deraadt 2024-08-18 19:58:35 +00:00
parent 84e87b2141
commit 0e2029b519
2 changed files with 5 additions and 5 deletions

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@ -10,13 +10,13 @@ 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
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
(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

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@ -1,4 +1,4 @@
/* $OpenBSD: cal.c,v 1.31 2022/12/04 23:50:47 cheloha Exp $ */
/* $OpenBSD: cal.c,v 1.32 2024/08/18 19:58:35 deraadt Exp $ */
/* $NetBSD: cal.c,v 1.6 1995/03/26 03:10:24 glass Exp $ */
/*
@ -261,7 +261,7 @@ week(int day, int month, int year)
shift = 1;
if (yearday < firstsunday)
return (1);
if (firstweekday > THURSDAY - 1)
if (firstweekday > THURSDAY - 1)
shift = 2;
return ((((yearday + 1) - (weekday - 1)) / 7) + shift);
}
@ -291,7 +291,7 @@ isoweek(int day, int month, int year)
return 53 - (g - s) / 5;
else if (n > 364 + s)
return 1;
else
else
return n/7 + 1;
}