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336 lines
9.2 KiB
C
336 lines
9.2 KiB
C
/*


* SPDXLicenseIdentifier: BSD2ClauseFreeBSD


*


* 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.


*/




#include <sys/cdefs.h>


#include <langinfo.h>


__FBSDID("$FreeBSD: head/lib/libcalendar/calendar.c 326219 20171126 02:00:33Z pfg $");




#include "calendar.h"


extern int weekstart;




#ifndef NULL


#define NULL 0


#endif




/*


* 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 15821005. This is the reverse of the function ndaysg().


*/


date *


gdate(int ndays, date *dt)


{


int const *montht; /* monthtable */


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 15821004 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 daysreturn 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 y11 */


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 sunday/monday is the answer),


* otherwise this week is week 1 and the last sunday/monday is the


* answer.


*/


/* 3 may or may not be correct, better use what the locale says */


if ((wd = weekday(nd) + 1  weekstart) >= *nl_langinfo(_NL_TIME_WEEK_1STWEEK))


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 19971117 */


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);


}
