/* * Copyright (C) 2012 Michael Brown . * * 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 2 of the * License, or 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, write to the Free Software * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA * 02110-1301, USA. * * You can also choose to distribute this program under the terms of * the Unmodified Binary Distribution Licence (as given in the file * COPYING.UBDL), provided that you have satisfied its requirements. */ FILE_LICENCE ( GPL2_OR_LATER_OR_UBDL ); #include /** @file * * Date and time * * POSIX:2008 section 4.15 defines "seconds since the Epoch" as an * abstract measure approximating the number of seconds that have * elapsed since the Epoch, excluding leap seconds. The formula given * is * * tm_sec + tm_min*60 + tm_hour*3600 + tm_yday*86400 + * (tm_year-70)*31536000 + ((tm_year-69)/4)*86400 - * ((tm_year-1)/100)*86400 + ((tm_year+299)/400)*86400 * * This calculation assumes that leap years occur in each year that is * either divisible by 4 but not divisible by 100, or is divisible by * 400. */ /** Days of week (for debugging) */ static const char *weekdays[] = { "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat" }; /** * Determine whether or not year is a leap year * * @v tm_year Years since 1900 * @v is_leap_year Year is a leap year */ static int is_leap_year ( int tm_year ) { int leap_year = 0; if ( ( tm_year % 4 ) == 0 ) leap_year = 1; if ( ( tm_year % 100 ) == 0 ) leap_year = 0; if ( ( tm_year % 400 ) == 100 ) leap_year = 1; return leap_year; } /** * Calculate number of leap years since 1900 * * @v tm_year Years since 1900 * @v num_leap_years Number of leap years */ static int leap_years_to_end ( int tm_year ) { int leap_years = 0; leap_years += ( tm_year / 4 ); leap_years -= ( tm_year / 100 ); leap_years += ( ( tm_year + 300 ) / 400 ); return leap_years; } /** * Calculate day of week * * @v tm_year Years since 1900 * @v tm_mon Month of year [0,11] * @v tm_day Day of month [1,31] */ static int day_of_week ( int tm_year, int tm_mon, int tm_mday ) { static const uint8_t offset[12] = { 1, 4, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5 }; int pseudo_year = tm_year; if ( tm_mon < 2 ) pseudo_year--; return ( ( pseudo_year + leap_years_to_end ( pseudo_year ) + offset[tm_mon] + tm_mday ) % 7 ); } /** Days from start of year until start of months (in non-leap years) */ static const uint16_t days_to_month_start[] = { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 }; /** * Calculate seconds since the Epoch * * @v tm Broken-down time * @ret time Seconds since the Epoch */ time_t mktime ( struct tm *tm ) { int days_since_epoch; int seconds_since_day; time_t seconds; /* Calculate day of year */ tm->tm_yday = ( ( tm->tm_mday - 1 ) + days_to_month_start[ tm->tm_mon ] ); if ( ( tm->tm_mon >= 2 ) && is_leap_year ( tm->tm_year ) ) tm->tm_yday++; /* Calculate day of week */ tm->tm_wday = day_of_week ( tm->tm_year, tm->tm_mon, tm->tm_mday ); /* Calculate seconds since the Epoch */ days_since_epoch = ( tm->tm_yday + ( 365 * tm->tm_year ) - 25567 + leap_years_to_end ( tm->tm_year - 1 ) ); seconds_since_day = ( ( ( ( tm->tm_hour * 60 ) + tm->tm_min ) * 60 ) + tm->tm_sec ); seconds = ( ( ( ( time_t ) days_since_epoch ) * ( ( time_t ) 86400 ) ) + seconds_since_day ); DBGC ( &weekdays, "TIME %04d-%02d-%02d %02d:%02d:%02d => %lld (%s, " "day %d)\n", ( tm->tm_year + 1900 ), ( tm->tm_mon + 1 ), tm->tm_mday, tm->tm_hour, tm->tm_min, tm->tm_sec, seconds, weekdays[ tm->tm_wday ], tm->tm_yday ); return seconds; }