xref: /dragonfly/lib/libcalendar/calendar.c (revision cfd1aba3)
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  * $FreeBSD: src/lib/libcalendar/calendar.c,v 1.3 1999/08/28 00:04:03 peter Exp $
27  */
28 
29 #include <sys/param.h>
30 #include "calendar.h"
31 
32 /*
33  * For each month tabulate the number of days elapsed in a year before the
34  * month. This assumes the internal date representation, where a year
35  * starts on March 1st. So we don't need a special table for leap years.
36  * But we do need a special table for the year 1582, since 10 days are
37  * deleted in October. This is month1s for the switch from Julian to
38  * Gregorian calendar.
39  */
40 static int const month1[] =
41     {0, 31, 61, 92, 122, 153, 184, 214, 245, 275, 306, 337};
42    /*  M   A   M   J    J    A    S    O    N    D    J */
43 static int const month1s[]=
44     {0, 31, 61, 92, 122, 153, 184, 214, 235, 265, 296, 327};
45 
46 typedef struct date date;
47 
48 /* The last day of Julian calendar, in internal and ndays representation */
49 static int nswitch;	/* The last day of Julian calendar */
50 static date jiswitch = {1582, 7, 3};
51 
52 static date	*date2idt(date *idt, date *dt);
53 static date	*idt2date(date *dt, date *idt);
54 static int	 ndaysji(date *idt);
55 static int	 ndaysgi(date *idt);
56 static int	 firstweek(int year);
57 
58 /*
59  * Compute the Julian date from the number of days elapsed since
60  * March 1st of year zero.
61  */
62 date *
63 jdate(int ndays, date *dt)
64 {
65 	date    idt;		/* Internal date representation */
66 	int     r;		/* hold the rest of days */
67 
68 	/*
69 	 * Compute the year by starting with an approximation not smaller
70 	 * than the answer and using linear search for the greatest
71 	 * year which does not begin after ndays.
72 	 */
73 	idt.y = ndays / 365;
74 	idt.m = 0;
75 	idt.d = 0;
76 	while ((r = ndaysji(&idt)) > ndays)
77 		idt.y--;
78 
79 	/*
80 	 * Set r to the days left in the year and compute the month by
81 	 * linear search as the largest month that does not begin after r
82 	 * days.
83 	 */
84 	r = ndays - r;
85 	for (idt.m = 11; month1[idt.m] > r; idt.m--)
86 		;
87 
88 	/* Compute the days left in the month */
89 	idt.d = r - month1[idt.m];
90 
91 	/* return external representation of the date */
92 	return (idt2date(dt, &idt));
93 }
94 
95 /*
96  * Return the number of days since March 1st of the year zero.
97  * The date is given according to Julian calendar.
98  */
99 int
100 ndaysj(date *dt)
101 {
102 	date    idt;		/* Internal date representation */
103 
104 	if (date2idt(&idt, dt) == NULL)
105 		return (-1);
106 	else
107 		return (ndaysji(&idt));
108 }
109 
110 /*
111  * Same as above, where the Julian date is given in internal notation.
112  * This formula shows the beauty of this notation.
113  */
114 static int
115 ndaysji(date * idt)
116 {
117 
118 	return (idt->d + month1[idt->m] + idt->y * 365 + idt->y / 4);
119 }
120 
121 /*
122  * Compute the date according to the Gregorian calendar from the number of
123  * days since March 1st, year zero. The date computed will be Julian if it
124  * is older than 1582-10-05. This is the reverse of the function ndaysg().
125  */
126 date   *
127 gdate(int ndays, date *dt)
128 {
129 	int const *montht;	/* month-table */
130 	date    idt;		/* for internal date representation */
131 	int     r;		/* holds the rest of days */
132 
133 	/*
134 	 * Compute the year by starting with an approximation not smaller
135 	 * than the answer and search linearly for the greatest year not
136 	 * starting after ndays.
137 	 */
138 	idt.y = ndays / 365;
139 	idt.m = 0;
140 	idt.d = 0;
141 	while ((r = ndaysgi(&idt)) > ndays)
142 		idt.y--;
143 
144 	/*
145 	 * Set ndays to the number of days left and compute by linear
146 	 * search the greatest month which does not start after ndays. We
147 	 * use the table month1 which provides for each month the number
148 	 * of days that elapsed in the year before that month. Here the
149 	 * year 1582 is special, as 10 days are left out in October to
150 	 * resynchronize the calendar with the earth's orbit. October 4th
151 	 * 1582 is followed by October 15th 1582. We use the "switch"
152 	 * table month1s for this year.
153 	 */
154 	ndays = ndays - r;
155 	if (idt.y == 1582)
156 		montht = month1s;
157 	else
158 		montht = month1;
159 
160 	for (idt.m = 11; montht[idt.m] > ndays; idt.m--)
161 		;
162 
163 	idt.d = ndays - montht[idt.m]; /* the rest is the day in month */
164 
165 	/* Advance ten days deleted from October if after switch in Oct 1582 */
166 	if (idt.y == jiswitch.y && idt.m == jiswitch.m && jiswitch.d < idt.d)
167 		idt.d += 10;
168 
169 	/* return external representation of found date */
170 	return (idt2date(dt, &idt));
171 }
172 
173 /*
174  * Return the number of days since March 1st of the year zero. The date is
175  * assumed Gregorian if younger than 1582-10-04 and Julian otherwise. This
176  * is the reverse of gdate.
177  */
178 int
179 ndaysg(date *dt)
180 {
181 	date    idt;		/* Internal date representation */
182 
183 	if (date2idt(&idt, dt) == NULL)
184 		return (-1);
185 	return (ndaysgi(&idt));
186 }
187 
188 /*
189  * Same as above, but with the Gregorian date given in internal
190  * representation.
191  */
192 static int
193 ndaysgi(date *idt)
194 {
195 	int     nd;		/* Number of days--return value */
196 
197 	/* Cache nswitch if not already done */
198 	if (nswitch == 0)
199 		nswitch = ndaysji(&jiswitch);
200 
201 	/*
202 	 * Assume Julian calendar and adapt to Gregorian if necessary, i. e.
203 	 * younger than nswitch. Gregori deleted
204 	 * the ten days from Oct 5th to Oct 14th 1582.
205 	 * Thereafter years which are multiples of 100 and not multiples
206 	 * of 400 were not leap years anymore.
207 	 * This makes the average length of a year
208 	 * 365d +.25d - .01d + .0025d = 365.2425d. But the tropical
209 	 * year measures 365.2422d. So in 10000/3 years we are
210 	 * again one day ahead of the earth. Sigh :-)
211 	 * (d is the average length of a day and tropical year is the
212 	 * time from one spring point to the next.)
213 	 */
214 	if ((nd = ndaysji(idt)) == -1)
215 		return (-1);
216 	if (idt->y >= 1600)
217 		nd = (nd - 10 - (idt->y - 1600) / 100 + (idt->y - 1600) / 400);
218 	else if (nd > nswitch)
219 		nd -= 10;
220 	return (nd);
221 }
222 
223 /*
224  * Compute the week number from the number of days since March 1st year 0.
225  * The weeks are numbered per year starting with 1. If the first
226  * week of a year includes at least four days of that year it is week 1,
227  * otherwise it gets the number of the last week of the previous year.
228  * The variable y will be filled with the year that contains the greater
229  * part of the week.
230  */
231 int
232 week(int nd, int *y)
233 {
234 	date    dt;
235 	int     fw;		/* 1st day of week 1 of previous, this and
236 				 * next year */
237 	gdate(nd, &dt);
238 	for (*y = dt.y + 1; nd < (fw = firstweek(*y)); (*y)--)
239 		;
240 	return ((nd - fw) / 7 + 1);
241 }
242 
243 /* return the first day of week 1 of year y */
244 static int
245 firstweek(int y)
246 {
247 	date idt;
248 	int nd, wd;
249 
250 	idt.y = y - 1;   /* internal representation of y-1-1 */
251 	idt.m = 10;
252 	idt.d = 0;
253 
254 	nd = ndaysgi(&idt);
255 	/*
256 	 * If more than 3 days of this week are in the preceding year, the
257 	 * next week is week 1 (and the next monday is the answer),
258 	 * otherwise this week is week 1 and the last monday is the
259 	 * answer.
260 	 */
261 	if ((wd = weekday(nd)) > 3)
262 		return (nd - wd + 7);
263 	else
264 		return (nd - wd);
265 }
266 
267 /* return the weekday (Mo = 0 .. Su = 6) */
268 int
269 weekday(int nd)
270 {
271 	date dmondaygi = {1997, 8, 16}; /* Internal repr. of 1997-11-17 */
272 	static int nmonday;             /* ... which is a monday        */
273 
274 	/* Cache the daynumber of one monday */
275 	if (nmonday == 0)
276 		nmonday = ndaysgi(&dmondaygi);
277 
278 	/* return (nd - nmonday) modulo 7 which is the weekday */
279 	nd = (nd - nmonday) % 7;
280 	if (nd < 0)
281 		return (nd + 7);
282 	else
283 		return (nd);
284 }
285 
286 /*
287  * Convert a date to internal date representation: The year starts on
288  * March 1st, month and day numbering start at zero. E. g. March 1st of
289  * year zero is written as y=0, m=0, d=0.
290  */
291 static date *
292 date2idt(date *idt, date *dt)
293 {
294 
295 	idt->d = dt->d - 1;
296 	if (dt->m > 2) {
297 		idt->m = dt->m - 3;
298 		idt->y = dt->y;
299 	} else {
300 		idt->m = dt->m + 9;
301 		idt->y = dt->y - 1;
302 	}
303 	if (idt->m < 0 || idt->m > 11 || idt->y < 0)
304 		return (NULL);
305 	else
306 		return idt;
307 }
308 
309 /* Reverse of date2idt */
310 static date *
311 idt2date(date *dt, date *idt)
312 {
313 
314 	dt->d = idt->d + 1;
315 	if (idt->m < 10) {
316 		dt->m = idt->m + 3;
317 		dt->y = idt->y;
318 	} else {
319 		dt->m = idt->m - 9;
320 		dt->y = idt->y + 1;
321 	}
322 	if (dt->m < 1)
323 		return (NULL);
324 	else
325 		return (dt);
326 }
327