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