/*
* Copyright (C) 2012 Michael Brown <mbrown@fensystems.co.uk>.
*
* 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.
*/
FILE_LICENCE ( GPL2_OR_LATER );
#include <time.h>
/** @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;
}
