1 /*
2 * Copyright (C) 2012 Michael Brown <mbrown@fensystems.co.uk>.
3 *
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.
8 *
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.
13 *
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
17 * 02110-1301, USA.
18 *
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.
22 */
23
24 FILE_LICENCE ( GPL2_OR_LATER_OR_UBDL );
25
26 #include <time.h>
27
28 /** @file
29 *
30 * Date and time
31 *
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
35 * is
36 *
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
40 *
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
43 * 400.
44 */
45
46 /** Current system clock offset */
47 signed long time_offset;
48
49 /** Days of week (for debugging) */
50 static const char *weekdays[] = {
51 "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"
52 };
53
54 /**
55 * Determine whether or not year is a leap year
56 *
57 * @v tm_year Years since 1900
58 * @v is_leap_year Year is a leap year
59 */
is_leap_year(int tm_year)60 static int is_leap_year ( int tm_year ) {
61 int leap_year = 0;
62
63 if ( ( tm_year % 4 ) == 0 )
64 leap_year = 1;
65 if ( ( tm_year % 100 ) == 0 )
66 leap_year = 0;
67 if ( ( tm_year % 400 ) == 100 )
68 leap_year = 1;
69
70 return leap_year;
71 }
72
73 /**
74 * Calculate number of leap years since 1900
75 *
76 * @v tm_year Years since 1900
77 * @v num_leap_years Number of leap years
78 */
leap_years_to_end(int tm_year)79 static int leap_years_to_end ( int tm_year ) {
80 int leap_years = 0;
81
82 leap_years += ( tm_year / 4 );
83 leap_years -= ( tm_year / 100 );
84 leap_years += ( ( tm_year + 300 ) / 400 );
85
86 return leap_years;
87 }
88
89 /**
90 * Calculate day of week
91 *
92 * @v tm_year Years since 1900
93 * @v tm_mon Month of year [0,11]
94 * @v tm_day Day of month [1,31]
95 */
day_of_week(int tm_year,int tm_mon,int tm_mday)96 static int day_of_week ( int tm_year, int tm_mon, int tm_mday ) {
97 static const uint8_t offset[12] =
98 { 1, 4, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5 };
99 int pseudo_year = tm_year;
100
101 if ( tm_mon < 2 )
102 pseudo_year--;
103 return ( ( pseudo_year + leap_years_to_end ( pseudo_year ) +
104 offset[tm_mon] + tm_mday ) % 7 );
105 }
106
107 /** Days from start of year until start of months (in non-leap years) */
108 static const uint16_t days_to_month_start[] =
109 { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
110
111 /**
112 * Calculate seconds since the Epoch
113 *
114 * @v tm Broken-down time
115 * @ret time Seconds since the Epoch
116 */
mktime(struct tm * tm)117 time_t mktime ( struct tm *tm ) {
118 int days_since_epoch;
119 int seconds_since_day;
120 time_t seconds;
121
122 /* Calculate day of year */
123 tm->tm_yday = ( ( tm->tm_mday - 1 ) +
124 days_to_month_start[ tm->tm_mon ] );
125 if ( ( tm->tm_mon >= 2 ) && is_leap_year ( tm->tm_year ) )
126 tm->tm_yday++;
127
128 /* Calculate day of week */
129 tm->tm_wday = day_of_week ( tm->tm_year, tm->tm_mon, tm->tm_mday );
130
131 /* Calculate seconds since the Epoch */
132 days_since_epoch = ( tm->tm_yday + ( 365 * tm->tm_year ) - 25567 +
133 leap_years_to_end ( tm->tm_year - 1 ) );
134 seconds_since_day =
135 ( ( ( ( tm->tm_hour * 60 ) + tm->tm_min ) * 60 ) + tm->tm_sec );
136 seconds = ( ( ( ( time_t ) days_since_epoch ) * ( ( time_t ) 86400 ) ) +
137 seconds_since_day );
138
139 DBGC ( &weekdays, "TIME %04d-%02d-%02d %02d:%02d:%02d => %lld (%s, "
140 "day %d)\n", ( tm->tm_year + 1900 ), ( tm->tm_mon + 1 ),
141 tm->tm_mday, tm->tm_hour, tm->tm_min, tm->tm_sec, seconds,
142 weekdays[ tm->tm_wday ], tm->tm_yday );
143
144 return seconds;
145 }
146