260ba0bff1
Mainly focus on files that use BSD 2-Clause license, however the tool I was using mis-identified many licenses so this was mostly a manual - error prone - task. The Software Package Data Exchange (SPDX) group provides a specification to make it easier for automated tools to detect and summarize well known opensource licenses. We are gradually adopting the specification, noting that the tags are considered only advisory and do not, in any way, superceed or replace the license texts.
333 lines
9.0 KiB
C
333 lines
9.0 KiB
C
/*-
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* SPDX-License-Identifier: BSD-2-Clause-FreeBSD
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*
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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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#include <sys/cdefs.h>
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__FBSDID("$FreeBSD$");
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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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