/*
* day_in_week
* return the 0 based day number for any date from 1 Jan. 1 to
* 31 Dec. 9999. Assumes the Gregorian reformation eliminates
* 3 Sep. 1752 through 13 Sep. 1752. Returns Thursday for all
* missing days.
*/
int
day_in_week(int day, int month, int year) {
long temp;
temp = (long)(year - 1) * 365 + leap_years_since_year_1(year - 1)
+ day_in_year(day, month, year);
if (temp < FIRST_MISSING_DAY)
return ((temp - 1 + SATURDAY) % 7);
if (temp >= (FIRST_MISSING_DAY + NUMBER_MISSING_DAYS))
return (((temp - 1 + SATURDAY) - NUMBER_MISSING_DAYS) % 7);
return (THURSDAY);
}
/*
* day_array --
* Fill in an array of 42 integers with a calendar. Assume for a moment
* that you took the (maximum) 6 rows in a calendar and stretched them
* out end to end. You would have 42 numbers or spaces. This routine
* builds that array for any month from Jan. 1 through Dec. 9999.
*/
static void day_array(unsigned month, unsigned year, unsigned *days)
{
unsigned long temp;
unsigned i;
unsigned day, dw, dm;
memset(days, SPACE, MAXDAYS * sizeof(int));
if ((month == 9) && (year == 1752)) {
/* Assumes the Gregorian reformation eliminates
* 3 Sep. 1752 through 13 Sep. 1752.
*/
unsigned j_offset = julian * 244;
size_t oday = 0;
do {
days[oday+2] = sep1752[oday] + j_offset;
} while (++oday < sizeof(sep1752));
return;
}
/* day_in_year
* return the 1 based day number within the year
*/
day = 1;
if ((month > 2) && leap_year(year)) {
++day;
}
i = month;
while (i) {
day += days_in_month[--i];
}
/* day_in_week
* return the 0 based day number for any date from 1 Jan. 1 to
* 31 Dec. 9999. Assumes the Gregorian reformation eliminates
* 3 Sep. 1752 through 13 Sep. 1752. Returns Thursday for all
* missing days.
*/
temp = (long)(year - 1) * 365 + leap_years_since_year_1(year - 1) + day;
if (temp < FIRST_MISSING_DAY) {
dw = ((temp - 1 + SATURDAY) % 7);
} else {
dw = (((temp - 1 + SATURDAY) - NUMBER_MISSING_DAYS) % 7);
}
if (!julian) {
day = 1;
}
dm = days_in_month[month];
if ((month == 2) && leap_year(year)) {
++dm;
}
do {
days[dw++] = day++;
} while (--dm);
}
/*
* day_array --
* Fill in an array of 42 integers with a calendar. Assume for a moment
* that you took the (maximum) 6 rows in a calendar and stretched them
* out end to end. You would have 42 numbers or spaces. This routine
* builds that array for any month from Jan. 1 through Dec. 9999.
*/
static void day_array(int month, int year, int *days)
{
long temp;
int i;
int j_offset;
int day, dw, dm;
memset(days, SPACE, MAXDAYS * sizeof(int));
if ((month == 9) && (year == 1752)) {
j_offset = julian * 244;
i = 0;
do {
days[i+2] = sep1752[i] + j_offset;
} while (++i < sizeof(sep1752));
return;
}
/* day_in_year
* return the 1 based day number within the year
*/
day = 1;
if ((month > 2) && leap_year(year)) {
++day;
}
i = month;
while (i) {
day += days_in_month[--i];
}
/* day_in_week
* return the 0 based day number for any date from 1 Jan. 1 to
* 31 Dec. 9999. Assumes the Gregorian reformation eliminates
* 3 Sep. 1752 through 13 Sep. 1752. Returns Thursday for all
* missing days.
*/
dw = THURSDAY;
temp = (long)(year - 1) * 365 + leap_years_since_year_1(year - 1)
+ day;
if (temp < FIRST_MISSING_DAY) {
dw = ((temp - 1 + SATURDAY) % 7);
} else if (temp >= (FIRST_MISSING_DAY + NUMBER_MISSING_DAYS)) {
dw = (((temp - 1 + SATURDAY) - NUMBER_MISSING_DAYS) % 7);
}
if (!julian) {
day = 1;
}
dm = days_in_month[month];
if ((month == 2) && leap_year(year)) {
++dm;
}
while (dm) {
days[dw++] = day++;
--dm;
}
}