2 * Copyright (C) 2012 Michael Brown <mbrown@fensystems.co.uk>.
4 * This program is free software; you can redistribute it and/or
5 * modify it under the terms of the GNU General Public License as
6 * published by the Free Software Foundation; either version 2 of the
7 * License, or any later version.
9 * This program is distributed in the hope that it will be useful, but
10 * WITHOUT ANY WARRANTY; without even the implied warranty of
11 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
12 * General Public License for more details.
14 * You should have received a copy of the GNU General Public License
15 * along with this program; if not, write to the Free Software
16 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA
19 * You can also choose to distribute this program under the terms of
20 * the Unmodified Binary Distribution Licence (as given in the file
21 * COPYING.UBDL), provided that you have satisfied its requirements.
24 FILE_LICENCE ( GPL2_OR_LATER_OR_UBDL
);
32 * POSIX:2008 section 4.15 defines "seconds since the Epoch" as an
33 * abstract measure approximating the number of seconds that have
34 * elapsed since the Epoch, excluding leap seconds. The formula given
37 * tm_sec + tm_min*60 + tm_hour*3600 + tm_yday*86400 +
38 * (tm_year-70)*31536000 + ((tm_year-69)/4)*86400 -
39 * ((tm_year-1)/100)*86400 + ((tm_year+299)/400)*86400
41 * This calculation assumes that leap years occur in each year that is
42 * either divisible by 4 but not divisible by 100, or is divisible by
46 /** Days of week (for debugging) */
47 static const char *weekdays
[] = {
48 "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"
52 * Determine whether or not year is a leap year
54 * @v tm_year Years since 1900
55 * @v is_leap_year Year is a leap year
57 static int is_leap_year ( int tm_year
) {
60 if ( ( tm_year
% 4 ) == 0 )
62 if ( ( tm_year
% 100 ) == 0 )
64 if ( ( tm_year
% 400 ) == 100 )
71 * Calculate number of leap years since 1900
73 * @v tm_year Years since 1900
74 * @v num_leap_years Number of leap years
76 static int leap_years_to_end ( int tm_year
) {
79 leap_years
+= ( tm_year
/ 4 );
80 leap_years
-= ( tm_year
/ 100 );
81 leap_years
+= ( ( tm_year
+ 300 ) / 400 );
87 * Calculate day of week
89 * @v tm_year Years since 1900
90 * @v tm_mon Month of year [0,11]
91 * @v tm_day Day of month [1,31]
93 static int day_of_week ( int tm_year
, int tm_mon
, int tm_mday
) {
94 static const uint8_t offset
[12] =
95 { 1, 4, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5 };
96 int pseudo_year
= tm_year
;
100 return ( ( pseudo_year
+ leap_years_to_end ( pseudo_year
) +
101 offset
[tm_mon
] + tm_mday
) % 7 );
104 /** Days from start of year until start of months (in non-leap years) */
105 static const uint16_t days_to_month_start
[] =
106 { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
109 * Calculate seconds since the Epoch
111 * @v tm Broken-down time
112 * @ret time Seconds since the Epoch
114 time_t mktime ( struct tm
*tm
) {
115 int days_since_epoch
;
116 int seconds_since_day
;
119 /* Calculate day of year */
120 tm
->tm_yday
= ( ( tm
->tm_mday
- 1 ) +
121 days_to_month_start
[ tm
->tm_mon
] );
122 if ( ( tm
->tm_mon
>= 2 ) && is_leap_year ( tm
->tm_year
) )
125 /* Calculate day of week */
126 tm
->tm_wday
= day_of_week ( tm
->tm_year
, tm
->tm_mon
, tm
->tm_mday
);
128 /* Calculate seconds since the Epoch */
129 days_since_epoch
= ( tm
->tm_yday
+ ( 365 * tm
->tm_year
) - 25567 +
130 leap_years_to_end ( tm
->tm_year
- 1 ) );
132 ( ( ( ( tm
->tm_hour
* 60 ) + tm
->tm_min
) * 60 ) + tm
->tm_sec
);
133 seconds
= ( ( ( ( time_t ) days_since_epoch
) * ( ( time_t ) 86400 ) ) +
136 DBGC ( &weekdays
, "TIME %04d-%02d-%02d %02d:%02d:%02d => %lld (%s, "
137 "day %d)\n", ( tm
->tm_year
+ 1900 ), ( tm
->tm_mon
+ 1 ),
138 tm
->tm_mday
, tm
->tm_hour
, tm
->tm_min
, tm
->tm_sec
, seconds
,
139 weekdays
[ tm
->tm_wday
], tm
->tm_yday
);