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+/*

+ * dayslib.c

+ *

+ * Small library to handle dates. The algorithms are mostly from Pascal's SWAG.

+ *

+ * History:

+ *       1/11/03 Creation from the old sources of ohdatabank.

+ *       8/01/04 Fix the yday2wday() function. Also generalize the formula

+ *               for a wider range of years (that is, pass the 2099 barrier).

+ *      15/09/06 Fix some comments, protect the macro GetYearDays.

+ *

+ * Author: Dario Rodriguez dario@softhome.net

+ * This code is dual licensed under the terms of the GNU LGPL/3-clause BSD.

+ */

+

+/* Constants */

+const int __NAcDaysNoLeap[13]=

+        {0,31,59,90,120,151,181,212,243,273,304,334,365};

+const int __NAcDaysLeap[13]=

+        {0,31,60,91,121,152,182,213,244,274,305,335,366};

+

+int *

+GetYearDays(int year)

+{

+	return(((((year)%4)==0 && !(((year)%100)==0)) \

+                || ((year)%400)==0)?__NAcDaysLeap:__NAcDaysNoLeap);

+}

+

+/* Function prototypes */

+int yday2mday(int year, int yday, int *month, int *day);

+int mday2yday(int year, int month, int day, int *yday);

+int yday2wday(int year, int yday, int *wday);

+

+/* Function bodies */

+/* yday2mday(): returns the day-in-month of the specified day-in-year */

+int

+yday2mday(int year, int yday, int *month, int *day)

+{

+        const int *NAcDays;

+        NAcDays=GetYearDays(year);

+        for(*month=1;yday>NAcDays[*month];(*month)++)

+                ;

+        *day=yday-NAcDays[(*month)-1];

+        return(year);

+}

+

+/* yday2mday(): returns the day-in-year of the specified day-in-month */

+int

+mday2yday(int year, int month, int day, int *yday)

+{

+        const int *NAcDays;

+        NAcDays=GetYearDays(year);

+        *yday=NAcDays[month-1]+day;

+        return(year);

+}

+

+/* yday2mday(): returns the weekday (0:sunday) of the specified day-in-year */

+/* Warning: formula only valid for 1900- */

+int

+yday2wday(int year, int yday, int *wday)

+{

+#if 0

+        /* Simplified formula, valid only 1901-2099 */

+        *wday=(((year+3)/4)+year+4+yday)%7;

+#else

+        /* Full-blown formula, valid from 1900 onwards */

+        /* (ok, it works for years earlier to 1900, but I'm not sure when */

+        /* Also, it varies from country to country when they adopted the */

+        /* current calendary... see your encyclopaedia for more info on this).*/

+        *wday=(((year+3)/4)+year+4+yday-((year-1)/100-19)+((year-1)/400-4))%7;

+#endif

+        return(year);

+}

+

+#ifdef UNIT_TESTING

+#include <stdio.h>

+int

+main(void)

+{

+        struct {

+                int day;

+                int month;

+                int year;

+                int wday;

+        } Tests[]={{1,1,2003,3},{31,12,2003,3},{1,1,2004,4},{1,2,2004,0},{1,3,2004,1},

+                   {1,1,2000,6},{1,1,2001,1},{1,1,2002,2},{1,1,2003,3},{1,1,2004,4},{1,1,2005,6},

+                   {1,1,2099,4},{1,1,2100,5},{1,1,2101,6},{1,1,2102,0},{1,7,2087,2},

+                   {1,1,1897,5},{1,1,1898,6},{1,1,1899,0},{1,1,1900,1},{1,1,1901,2},

+                   {12,10,2008,0},{0}};

+        long i;

+        int day,month,year,wday;

+        int resyday,reswday;

+        for(i=0;Tests[i].day>=1;i++) {

+                day=Tests[i].day,month=Tests[i].month;

+                year=Tests[i].year,wday=Tests[i].wday;

+                mday2yday(year,month,day,&resyday);

+                yday2wday(year,resyday,&reswday);

+                printf("%2i/%02i/%04i(%i) -> yday:%i,wday:%i... %s\n",day,month,year,wday,resyday,reswday,(wday!=reswday)?"ERROR":"Ok");

+        }

+        return(0);

+}

+#endif