calendar.c revision 331722
1/*-
2 * Copyright (c) 1997 Wolfgang Helbig
3 * All rights reserved.
4 *
5 * Redistribution and use in source and binary forms, with or without
6 * modification, are permitted provided that the following conditions
7 * are met:
8 * 1. Redistributions of source code must retain the above copyright
9 *    notice, this list of conditions and the following disclaimer.
10 * 2. Redistributions in binary form must reproduce the above copyright
11 *    notice, this list of conditions and the following disclaimer in the
12 *    documentation and/or other materials provided with the distribution.
13 *
14 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
15 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
16 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
17 * ARE DISCLAIMED.  IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
18 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
19 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
20 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
21 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
22 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
23 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
24 * SUCH DAMAGE.
25 */
26
27#include <sys/cdefs.h>
28__FBSDID("$FreeBSD: stable/11/lib/libcalendar/calendar.c 331722 2018-03-29 02:50:57Z eadler $");
29
30#include "calendar.h"
31
32#ifndef NULL
33#define NULL 0
34#endif
35
36/*
37 * For each month tabulate the number of days elapsed in a year before the
38 * month. This assumes the internal date representation, where a year
39 * starts on March 1st. So we don't need a special table for leap years.
40 * But we do need a special table for the year 1582, since 10 days are
41 * deleted in October. This is month1s for the switch from Julian to
42 * Gregorian calendar.
43 */
44static int const month1[] =
45    {0, 31, 61, 92, 122, 153, 184, 214, 245, 275, 306, 337};
46   /*  M   A   M   J    J    A    S    O    N    D    J */
47static int const month1s[]=
48    {0, 31, 61, 92, 122, 153, 184, 214, 235, 265, 296, 327};
49
50typedef struct date date;
51
52/* The last day of Julian calendar, in internal and ndays representation */
53static int nswitch;	/* The last day of Julian calendar */
54static date jiswitch = {1582, 7, 3};
55
56static date	*date2idt(date *idt, date *dt);
57static date	*idt2date(date *dt, date *idt);
58static int	 ndaysji(date *idt);
59static int	 ndaysgi(date *idt);
60static int	 firstweek(int year);
61
62/*
63 * Compute the Julian date from the number of days elapsed since
64 * March 1st of year zero.
65 */
66date *
67jdate(int ndays, date *dt)
68{
69	date    idt;		/* Internal date representation */
70	int     r;		/* hold the rest of days */
71
72	/*
73	 * Compute the year by starting with an approximation not smaller
74	 * than the answer and using linear search for the greatest
75	 * year which does not begin after ndays.
76	 */
77	idt.y = ndays / 365;
78	idt.m = 0;
79	idt.d = 0;
80	while ((r = ndaysji(&idt)) > ndays)
81		idt.y--;
82
83	/*
84	 * Set r to the days left in the year and compute the month by
85	 * linear search as the largest month that does not begin after r
86	 * days.
87	 */
88	r = ndays - r;
89	for (idt.m = 11; month1[idt.m] > r; idt.m--)
90		;
91
92	/* Compute the days left in the month */
93	idt.d = r - month1[idt.m];
94
95	/* return external representation of the date */
96	return (idt2date(dt, &idt));
97}
98
99/*
100 * Return the number of days since March 1st of the year zero.
101 * The date is given according to Julian calendar.
102 */
103int
104ndaysj(date *dt)
105{
106	date    idt;		/* Internal date representation */
107
108	if (date2idt(&idt, dt) == NULL)
109		return (-1);
110	else
111		return (ndaysji(&idt));
112}
113
114/*
115 * Same as above, where the Julian date is given in internal notation.
116 * This formula shows the beauty of this notation.
117 */
118static int
119ndaysji(date * idt)
120{
121
122	return (idt->d + month1[idt->m] + idt->y * 365 + idt->y / 4);
123}
124
125/*
126 * Compute the date according to the Gregorian calendar from the number of
127 * days since March 1st, year zero. The date computed will be Julian if it
128 * is older than 1582-10-05. This is the reverse of the function ndaysg().
129 */
130date   *
131gdate(int ndays, date *dt)
132{
133	int const *montht;	/* month-table */
134	date    idt;		/* for internal date representation */
135	int     r;		/* holds the rest of days */
136
137	/*
138	 * Compute the year by starting with an approximation not smaller
139	 * than the answer and search linearly for the greatest year not
140	 * starting after ndays.
141	 */
142	idt.y = ndays / 365;
143	idt.m = 0;
144	idt.d = 0;
145	while ((r = ndaysgi(&idt)) > ndays)
146		idt.y--;
147
148	/*
149	 * Set ndays to the number of days left and compute by linear
150	 * search the greatest month which does not start after ndays. We
151	 * use the table month1 which provides for each month the number
152	 * of days that elapsed in the year before that month. Here the
153	 * year 1582 is special, as 10 days are left out in October to
154	 * resynchronize the calendar with the earth's orbit. October 4th
155	 * 1582 is followed by October 15th 1582. We use the "switch"
156	 * table month1s for this year.
157	 */
158	ndays = ndays - r;
159	if (idt.y == 1582)
160		montht = month1s;
161	else
162		montht = month1;
163
164	for (idt.m = 11; montht[idt.m] > ndays; idt.m--)
165		;
166
167	idt.d = ndays - montht[idt.m]; /* the rest is the day in month */
168
169	/* Advance ten days deleted from October if after switch in Oct 1582 */
170	if (idt.y == jiswitch.y && idt.m == jiswitch.m && jiswitch.d < idt.d)
171		idt.d += 10;
172
173	/* return external representation of found date */
174	return (idt2date(dt, &idt));
175}
176
177/*
178 * Return the number of days since March 1st of the year zero. The date is
179 * assumed Gregorian if younger than 1582-10-04 and Julian otherwise. This
180 * is the reverse of gdate.
181 */
182int
183ndaysg(date *dt)
184{
185	date    idt;		/* Internal date representation */
186
187	if (date2idt(&idt, dt) == NULL)
188		return (-1);
189	return (ndaysgi(&idt));
190}
191
192/*
193 * Same as above, but with the Gregorian date given in internal
194 * representation.
195 */
196static int
197ndaysgi(date *idt)
198{
199	int     nd;		/* Number of days--return value */
200
201	/* Cache nswitch if not already done */
202	if (nswitch == 0)
203		nswitch = ndaysji(&jiswitch);
204
205	/*
206	 * Assume Julian calendar and adapt to Gregorian if necessary, i. e.
207	 * younger than nswitch. Gregori deleted
208	 * the ten days from Oct 5th to Oct 14th 1582.
209	 * Thereafter years which are multiples of 100 and not multiples
210	 * of 400 were not leap years anymore.
211	 * This makes the average length of a year
212	 * 365d +.25d - .01d + .0025d = 365.2425d. But the tropical
213	 * year measures 365.2422d. So in 10000/3 years we are
214	 * again one day ahead of the earth. Sigh :-)
215	 * (d is the average length of a day and tropical year is the
216	 * time from one spring point to the next.)
217	 */
218	if ((nd = ndaysji(idt)) == -1)
219		return (-1);
220	if (idt->y >= 1600)
221		nd = (nd - 10 - (idt->y - 1600) / 100 + (idt->y - 1600) / 400);
222	else if (nd > nswitch)
223		nd -= 10;
224	return (nd);
225}
226
227/*
228 * Compute the week number from the number of days since March 1st year 0.
229 * The weeks are numbered per year starting with 1. If the first
230 * week of a year includes at least four days of that year it is week 1,
231 * otherwise it gets the number of the last week of the previous year.
232 * The variable y will be filled with the year that contains the greater
233 * part of the week.
234 */
235int
236week(int nd, int *y)
237{
238	date    dt;
239	int     fw;		/* 1st day of week 1 of previous, this and
240				 * next year */
241	gdate(nd, &dt);
242	for (*y = dt.y + 1; nd < (fw = firstweek(*y)); (*y)--)
243		;
244	return ((nd - fw) / 7 + 1);
245}
246
247/* return the first day of week 1 of year y */
248static int
249firstweek(int y)
250{
251	date idt;
252	int nd, wd;
253
254	idt.y = y - 1;   /* internal representation of y-1-1 */
255	idt.m = 10;
256	idt.d = 0;
257
258	nd = ndaysgi(&idt);
259	/*
260	 * If more than 3 days of this week are in the preceding year, the
261	 * next week is week 1 (and the next monday is the answer),
262	 * otherwise this week is week 1 and the last monday is the
263	 * answer.
264	 */
265	if ((wd = weekday(nd)) > 3)
266		return (nd - wd + 7);
267	else
268		return (nd - wd);
269}
270
271/* return the weekday (Mo = 0 .. Su = 6) */
272int
273weekday(int nd)
274{
275	date dmondaygi = {1997, 8, 16}; /* Internal repr. of 1997-11-17 */
276	static int nmonday;             /* ... which is a monday        */
277
278	/* Cache the daynumber of one monday */
279	if (nmonday == 0)
280		nmonday = ndaysgi(&dmondaygi);
281
282	/* return (nd - nmonday) modulo 7 which is the weekday */
283	nd = (nd - nmonday) % 7;
284	if (nd < 0)
285		return (nd + 7);
286	else
287		return (nd);
288}
289
290/*
291 * Convert a date to internal date representation: The year starts on
292 * March 1st, month and day numbering start at zero. E. g. March 1st of
293 * year zero is written as y=0, m=0, d=0.
294 */
295static date *
296date2idt(date *idt, date *dt)
297{
298
299	idt->d = dt->d - 1;
300	if (dt->m > 2) {
301		idt->m = dt->m - 3;
302		idt->y = dt->y;
303	} else {
304		idt->m = dt->m + 9;
305		idt->y = dt->y - 1;
306	}
307	if (idt->m < 0 || idt->m > 11 || idt->y < 0)
308		return (NULL);
309	else
310		return idt;
311}
312
313/* Reverse of date2idt */
314static date *
315idt2date(date *dt, date *idt)
316{
317
318	dt->d = idt->d + 1;
319	if (idt->m < 10) {
320		dt->m = idt->m + 3;
321		dt->y = idt->y;
322	} else {
323		dt->m = idt->m - 9;
324		dt->y = idt->y + 1;
325	}
326	if (dt->m < 1)
327		return (NULL);
328	else
329		return (dt);
330}
331