/*-
* Copyright (c) 1997 Wolfgang Helbig
* All rights reserved.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
*
* THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*
* $FreeBSD: src/lib/libcalendar/calendar.c,v 1.3 1999/08/28 00:04:03 peter Exp $
*/
#include
#include "calendar.h"
/*
* For each month tabulate the number of days elapsed in a year before the
* month. This assumes the internal date representation, where a year
* starts on March 1st. So we don't need a special table for leap years.
* But we do need a special table for the year 1582, since 10 days are
* deleted in October. This is month1s for the switch from Julian to
* Gregorian calendar.
*/
static int const month1[] =
{0, 31, 61, 92, 122, 153, 184, 214, 245, 275, 306, 337};
/* M A M J J A S O N D J */
static int const month1s[]=
{0, 31, 61, 92, 122, 153, 184, 214, 235, 265, 296, 327};
typedef struct date date;
/* The last day of Julian calendar, in internal and ndays representation */
static int nswitch; /* The last day of Julian calendar */
static date jiswitch = {1582, 7, 3};
static date *date2idt(date *idt, date *dt);
static date *idt2date(date *dt, date *idt);
static int ndaysji(date *idt);
static int ndaysgi(date *idt);
static int firstweek(int year);
/*
* Compute the Julian date from the number of days elapsed since
* March 1st of year zero.
*/
date *
jdate(int ndays, date *dt)
{
date idt; /* Internal date representation */
int r; /* hold the rest of days */
/*
* Compute the year by starting with an approximation not smaller
* than the answer and using linear search for the greatest
* year which does not begin after ndays.
*/
idt.y = ndays / 365;
idt.m = 0;
idt.d = 0;
while ((r = ndaysji(&idt)) > ndays)
idt.y--;
/*
* Set r to the days left in the year and compute the month by
* linear search as the largest month that does not begin after r
* days.
*/
r = ndays - r;
for (idt.m = 11; month1[idt.m] > r; idt.m--)
;
/* Compute the days left in the month */
idt.d = r - month1[idt.m];
/* return external representation of the date */
return (idt2date(dt, &idt));
}
/*
* Return the number of days since March 1st of the year zero.
* The date is given according to Julian calendar.
*/
int
ndaysj(date *dt)
{
date idt; /* Internal date representation */
if (date2idt(&idt, dt) == NULL)
return (-1);
else
return (ndaysji(&idt));
}
/*
* Same as above, where the Julian date is given in internal notation.
* This formula shows the beauty of this notation.
*/
static int
ndaysji(date * idt)
{
return (idt->d + month1[idt->m] + idt->y * 365 + idt->y / 4);
}
/*
* Compute the date according to the Gregorian calendar from the number of
* days since March 1st, year zero. The date computed will be Julian if it
* is older than 1582-10-05. This is the reverse of the function ndaysg().
*/
date *
gdate(int ndays, date *dt)
{
int const *montht; /* month-table */
date idt; /* for internal date representation */
int r; /* holds the rest of days */
/*
* Compute the year by starting with an approximation not smaller
* than the answer and search linearly for the greatest year not
* starting after ndays.
*/
idt.y = ndays / 365;
idt.m = 0;
idt.d = 0;
while ((r = ndaysgi(&idt)) > ndays)
idt.y--;
/*
* Set ndays to the number of days left and compute by linear
* search the greatest month which does not start after ndays. We
* use the table month1 which provides for each month the number
* of days that elapsed in the year before that month. Here the
* year 1582 is special, as 10 days are left out in October to
* resynchronize the calendar with the earth's orbit. October 4th
* 1582 is followed by October 15th 1582. We use the "switch"
* table month1s for this year.
*/
ndays = ndays - r;
if (idt.y == 1582)
montht = month1s;
else
montht = month1;
for (idt.m = 11; montht[idt.m] > ndays; idt.m--)
;
idt.d = ndays - montht[idt.m]; /* the rest is the day in month */
/* Advance ten days deleted from October if after switch in Oct 1582 */
if (idt.y == jiswitch.y && idt.m == jiswitch.m && jiswitch.d < idt.d)
idt.d += 10;
/* return external representation of found date */
return (idt2date(dt, &idt));
}
/*
* Return the number of days since March 1st of the year zero. The date is
* assumed Gregorian if younger than 1582-10-04 and Julian otherwise. This
* is the reverse of gdate.
*/
int
ndaysg(date *dt)
{
date idt; /* Internal date representation */
if (date2idt(&idt, dt) == NULL)
return (-1);
return (ndaysgi(&idt));
}
/*
* Same as above, but with the Gregorian date given in internal
* representation.
*/
static int
ndaysgi(date *idt)
{
int nd; /* Number of days--return value */
/* Cache nswitch if not already done */
if (nswitch == 0)
nswitch = ndaysji(&jiswitch);
/*
* Assume Julian calendar and adapt to Gregorian if necessary, i. e.
* younger than nswitch. Gregori deleted
* the ten days from Oct 5th to Oct 14th 1582.
* Thereafter years which are multiples of 100 and not multiples
* of 400 were not leap years anymore.
* This makes the average length of a year
* 365d +.25d - .01d + .0025d = 365.2425d. But the tropical
* year measures 365.2422d. So in 10000/3 years we are
* again one day ahead of the earth. Sigh :-)
* (d is the average length of a day and tropical year is the
* time from one spring point to the next.)
*/
if ((nd = ndaysji(idt)) == -1)
return (-1);
if (idt->y >= 1600)
nd = (nd - 10 - (idt->y - 1600) / 100 + (idt->y - 1600) / 400);
else if (nd > nswitch)
nd -= 10;
return (nd);
}
/*
* Compute the week number from the number of days since March 1st year 0.
* The weeks are numbered per year starting with 1. If the first
* week of a year includes at least four days of that year it is week 1,
* otherwise it gets the number of the last week of the previous year.
* The variable y will be filled with the year that contains the greater
* part of the week.
*/
int
week(int nd, int *y)
{
date dt;
int fw; /* 1st day of week 1 of previous, this and
* next year */
gdate(nd, &dt);
for (*y = dt.y + 1; nd < (fw = firstweek(*y)); (*y)--)
;
return ((nd - fw) / 7 + 1);
}
/* return the first day of week 1 of year y */
static int
firstweek(int y)
{
date idt;
int nd, wd;
idt.y = y - 1; /* internal representation of y-1-1 */
idt.m = 10;
idt.d = 0;
nd = ndaysgi(&idt);
/*
* If more than 3 days of this week are in the preceding year, the
* next week is week 1 (and the next monday is the answer),
* otherwise this week is week 1 and the last monday is the
* answer.
*/
if ((wd = weekday(nd)) > 3)
return (nd - wd + 7);
else
return (nd - wd);
}
/* return the weekday (Mo = 0 .. Su = 6) */
int
weekday(int nd)
{
date dmondaygi = {1997, 8, 16}; /* Internal repr. of 1997-11-17 */
static int nmonday; /* ... which is a monday */
/* Cache the daynumber of one monday */
if (nmonday == 0)
nmonday = ndaysgi(&dmondaygi);
/* return (nd - nmonday) modulo 7 which is the weekday */
nd = (nd - nmonday) % 7;
if (nd < 0)
return (nd + 7);
else
return (nd);
}
/*
* Convert a date to internal date representation: The year starts on
* March 1st, month and day numbering start at zero. E. g. March 1st of
* year zero is written as y=0, m=0, d=0.
*/
static date *
date2idt(date *idt, date *dt)
{
idt->d = dt->d - 1;
if (dt->m > 2) {
idt->m = dt->m - 3;
idt->y = dt->y;
} else {
idt->m = dt->m + 9;
idt->y = dt->y - 1;
}
if (idt->m < 0 || idt->m > 11 || idt->y < 0)
return (NULL);
else
return idt;
}
/* Reverse of date2idt */
static date *
idt2date(date *dt, date *idt)
{
dt->d = idt->d + 1;
if (idt->m < 10) {
dt->m = idt->m + 3;
dt->y = idt->y;
} else {
dt->m = idt->m - 9;
dt->y = idt->y + 1;
}
if (dt->m < 1)
return (NULL);
else
return (dt);
}