fa18377023
In the man page Use ".Pp" instead of blank lines, adopt English and stress that the Julian->Gregorian switch took place at different dates in different countries. Suggested by: Garrett.
330 lines
8.9 KiB
C
330 lines
8.9 KiB
C
/*-
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* Copyright (c) 1997 Wolfgang Helbig
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* All rights reserved.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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*
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* THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*
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* $Id: calendar.c,v 1.1.1.1 1997/12/04 10:41:49 helbig Exp $
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*/
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#include "calendar.h"
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#ifndef NULL
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#define NULL 0
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#endif
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/*
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* For each month tabulate the number of days elapsed in a year before the
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* month. This assumes the internal date representation, where a year
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* starts on March 1st. So we don't need a special table for leap years.
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* But we do need a special table for the year 1582, since 10 days are
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* deleted in October. This is month1s for the switch from Julian to
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* Gregorian calendar.
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*/
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static int const month1[] =
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{0, 31, 61, 92, 122, 153, 184, 214, 245, 275, 306, 337};
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/* M A M J J A S O N D J */
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static int const month1s[]=
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{0, 31, 61, 92, 122, 153, 184, 214, 235, 265, 296, 327};
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typedef struct date date;
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/* The last day of Julian calendar, in internal and ndays representation */
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static int nswitch; /* The last day of Julian calendar */
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static date jiswitch = {1582, 7, 3};
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static date *date2idt(date *idt, date *dt);
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static date *idt2date(date *dt, date *idt);
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static int ndaysji(date *idt);
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static int ndaysgi(date *idt);
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static int firstweek(int year);
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/*
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* Compute the Julian date from the number of days elapsed since
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* March 1st of year zero.
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*/
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date *
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jdate(int ndays, date *dt)
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{
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date idt; /* Internal date representation */
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int r; /* hold the rest of days */
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/*
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* Compute the year by starting with an approximation not smaller
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* than the answer and using linear search for the greatest
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* year which does not begin after ndays.
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*/
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idt.y = ndays / 365;
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idt.m = 0;
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idt.d = 0;
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while ((r = ndaysji(&idt)) > ndays)
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idt.y--;
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/*
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* Set r to the days left in the year and compute the month by
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* linear search as the largest month that does not begin after r
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* days.
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*/
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r = ndays - r;
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for (idt.m = 11; month1[idt.m] > r; idt.m--)
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;
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/* Compute the days left in the month */
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idt.d = r - month1[idt.m];
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/* return external representation of the date */
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return (idt2date(dt, &idt));
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}
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/*
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* Return the number of days since March 1st of the year zero.
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* The date is given according to Julian calendar.
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*/
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int
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ndaysj(date *dt)
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{
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date idt; /* Internal date representation */
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if (date2idt(&idt, dt) == NULL)
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return (-1);
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else
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return (ndaysji(&idt));
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}
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/*
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* Same as above, where the Julian date is given in internal notation.
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* This formula shows the beauty of this notation.
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*/
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static int
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ndaysji(date * idt)
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{
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return (idt->d + month1[idt->m] + idt->y * 365 + idt->y / 4);
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}
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/*
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* Compute the date according to the Gregorian calendar from the number of
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* days since March 1st, year zero. The date computed will be Julian if it
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* is older than 1582-10-05. This is the reverse of the function ndaysg().
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*/
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date *
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gdate(int ndays, date *dt)
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{
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int const *montht; /* month-table */
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date idt; /* for internal date representation */
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int r; /* holds the rest of days */
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/*
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* Compute the year by starting with an approximation not smaller
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* than the answer and search linearly for the greatest year not
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* starting after ndays.
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*/
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idt.y = ndays / 365;
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idt.m = 0;
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idt.d = 0;
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while ((r = ndaysgi(&idt)) > ndays)
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idt.y--;
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/*
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* Set ndays to the number of days left and compute by linear
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* search the greatest month which does not start after ndays. We
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* use the table month1 which provides for each month the number
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* of days that elapsed in the year before that month. Here the
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* year 1582 is special, as 10 days are left out in October to
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* resynchronize the calendar with the earth's orbit. October 4th
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* 1582 is followed by October 15th 1582. We use the "switch"
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* table month1s for this year.
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*/
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ndays = ndays - r;
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if (idt.y == 1582)
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montht = month1s;
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else
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montht = month1;
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for (idt.m = 11; montht[idt.m] > ndays; idt.m--)
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;
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idt.d = ndays - montht[idt.m]; /* the rest is the day in month */
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/* Advance ten days deleted from October if after switch in Oct 1582 */
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if (idt.y == jiswitch.y && idt.m == jiswitch.m && jiswitch.d < idt.d)
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idt.d += 10;
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/* return external representation of found date */
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return (idt2date(dt, &idt));
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}
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/*
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* Return the number of days since March 1st of the year zero. The date is
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* assumed Gregorian if younger than 1582-10-04 and Julian otherwise. This
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* is the reverse of gdate.
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*/
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int
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ndaysg(date *dt)
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{
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date idt; /* Internal date representation */
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if (date2idt(&idt, dt) == NULL)
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return (-1);
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return (ndaysgi(&idt));
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}
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/*
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* Same as above, but with the Gregorian date given in internal
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* representation.
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*/
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static int
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ndaysgi(date *idt)
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{
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int nd; /* Number of days--return value */
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/* Cache nswitch if not already done */
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if (nswitch == 0)
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nswitch = ndaysji(&jiswitch);
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/*
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* Assume Julian calendar and adapt to Gregorian if necessary, i. e.
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* younger than nswitch. Gregori deleted
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* the ten days from Oct 5th to Oct 14th 1582.
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* Thereafter years which are multiples of 100 and not multiples
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* of 400 were not leap years anymore.
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* This makes the average length of a year
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* 365d +.25d - .01d + .0025d = 365.2425d. But the tropical
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* year measures 365.2422d. So in 10000/3 years we are
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* again one day ahead of the earth. Sigh :-)
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* (d is the average length of a day and tropical year is the
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* time from one spring point to the next.)
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*/
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if ((nd = ndaysji(idt)) == -1)
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return (-1);
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if (idt->y >= 1600)
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nd = (nd - 10 - (idt->y - 1600) / 100 + (idt->y - 1600) / 400);
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else if (nd > nswitch)
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nd -= 10;
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return (nd);
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}
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/*
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* Compute the week number from the number of days since March 1st year 0.
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* The weeks are numbered per year starting with 1. If the first
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* week of a year includes at least four days of that year it is week 1,
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* otherwise it gets the number of the last week of the previous year.
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* The variable y will be filled with the year that contains the greater
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* part of the week.
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*/
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int
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week(int nd, int *y)
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{
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date dt;
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int fw; /* 1st day of week 1 of previous, this and
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* next year */
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gdate(nd, &dt);
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for (*y = dt.y + 1; nd < (fw = firstweek(*y)); (*y)--)
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;
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return ((nd - fw) / 7 + 1);
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}
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/* return the first day of week 1 of year y */
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static int
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firstweek(int y)
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{
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date idt;
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int nd, wd;
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idt.y = y - 1; /* internal representation of y-1-1 */
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idt.m = 10;
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idt.d = 0;
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nd = ndaysgi(&idt);
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/*
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* If more than 3 days of this week are in the preceding year, the
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* next week is week 1 (and the next monday is the answer),
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* otherwise this week is week 1 and the last monday is the
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* answer.
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*/
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if ((wd = weekday(nd)) > 3)
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return (nd - wd + 7);
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else
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return (nd - wd);
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}
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/* return the weekday (Mo = 0 .. Su = 6) */
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int
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weekday(int nd)
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{
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date dmondaygi = {1997, 8, 16}; /* Internal repr. of 1997-11-17 */
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static int nmonday; /* ... which is a monday */
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/* Cache the daynumber of one monday */
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if (nmonday == 0)
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nmonday = ndaysgi(&dmondaygi);
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/* return (nd - nmonday) modulo 7 which is the weekday */
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nd = (nd - nmonday) % 7;
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if (nd < 0)
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return (nd + 7);
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else
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return (nd);
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}
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/*
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* Convert a date to internal date representation: The year starts on
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* March 1st, month and day numbering start at zero. E. g. March 1st of
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* year zero is written as y=0, m=0, d=0.
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*/
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static date *
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date2idt(date *idt, date *dt)
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{
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idt->d = dt->d - 1;
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if (dt->m > 2) {
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idt->m = dt->m - 3;
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idt->y = dt->y;
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} else {
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idt->m = dt->m + 9;
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idt->y = dt->y - 1;
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}
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if (idt->m < 0 || idt->m > 11 || idt->y < 0)
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return (NULL);
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else
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return idt;
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}
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/* Reverse of date2idt */
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static date *
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idt2date(date *dt, date *idt)
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{
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dt->d = idt->d + 1;
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if (idt->m < 10) {
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dt->m = idt->m + 3;
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dt->y = idt->y;
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} else {
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dt->m = idt->m - 9;
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dt->y = idt->y + 1;
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}
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if (dt->m < 1)
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return (NULL);
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else
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return (dt);
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}
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