timeutils.c

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/*
 * This program is free software; you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation; either version 3 of the License, or
 * (at your option) any later version.
 *
 * This program is distributed in the hope that it will be useful, but
 * WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
 * or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
 * for more details.
 *
 * You should have received a copy of the GNU General Public License along
 * with this program; if not, see <http://www.gnu.org/licenses/>
 */

#include "timeutils.h"

/* Test for leap year. Nonzero if YEAR is a leap year (every 4 years,
 except every 100th isn't, and every 400th is). */
# define __isleap(year) \
 ((year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0))

/* How many days come before each month (0-12).  */
static const uint16_t __mon_yday[2][13] = {
        /* Normal years.  */
        {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
        /* Leap years.  */
        {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
};

#define SECS_PER_HOUR   (60 * 60)
#define SECS_PER_DAY    (SECS_PER_HOUR * 24)

void date_from_timestamp(uint32_t timestamp, DateTimeT *date_time)
{
        uint16_t y;
        int16_t days;
        uint32_t rem;

        days = timestamp / SECS_PER_DAY;
        rem = timestamp % SECS_PER_DAY;

        date_time->hour = rem / SECS_PER_HOUR;
        rem %= SECS_PER_HOUR;
        date_time->min = rem / 60;
        date_time->sec = rem % 60;
        /* January 1, 1970 was a Thursday.  */
        date_time->wday = (4 + days) % 7;
        y = 1970;

#define DIV(a, b) ((a) / (b) - ((a) % (b) < 0))
#define LEAPS_THRU_END_OF(y) (DIV (y, 4) - DIV (y, 100) + DIV (y, 400))

        while (days < 0 || days >= (__isleap (y) ? 366 : 365))
        {
                /* Guess a corrected year, assuming 365 days per year.  */
                uint16_t yg = y + days / 365 - (days % 365 < 0);

                /* Adjust DAYS and Y to match the guessed year.  */
                days -= ((yg - y) * 365
                                 + LEAPS_THRU_END_OF (yg - 1)
                                 - LEAPS_THRU_END_OF (y - 1));
                y = yg;
        }
        date_time->year = y - 1900;
        if (date_time->year != y - 1900)
        {
                /* The year cannot be represented due to overflow.  */
                return;
        }
        const uint16_t *ip = __mon_yday[__isleap(y)];
        for (y = 11; days < (uint16_t)ip[y]; --y){
                continue;
        }
        days -= ip[y];
        date_time->mon = y;
        date_time->mday = days + 1;
}