1 /*
2 * This file is part of the MicroPython project, http://micropython.org/
3 *
4 * The MIT License (MIT)
5 *
6 * Copyright (c) 2013, 2014 Damien P. George
7 * Copyright (c) 2015 Daniel Campora
8 *
9 * Permission is hereby granted, free of charge, to any person obtaining a copy
10 * of this software and associated documentation files (the "Software"), to deal
11 * in the Software without restriction, including without limitation the rights
12 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
13 * copies of the Software, and to permit persons to whom the Software is
14 * furnished to do so, subject to the following conditions:
15 *
16 * The above copyright notice and this permission notice shall be included in
17 * all copies or substantial portions of the Software.
18 *
19 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
20 * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
21 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
22 * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
23 * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
24 * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
25 * THE SOFTWARE.
26 */
27
28 #include "py/obj.h"
29
30 #include "shared/timeutils/timeutils.h"
31
32 // LEAPOCH corresponds to 2000-03-01, which is a mod-400 year, immediately
33 // after Feb 29. We calculate seconds as a signed integer relative to that.
34 //
35 // Our timebase is relative to 2000-01-01.
36
37 #define LEAPOCH ((31 + 29) * 86400)
38
39 #define DAYS_PER_400Y (365 * 400 + 97)
40 #define DAYS_PER_100Y (365 * 100 + 24)
41 #define DAYS_PER_4Y (365 * 4 + 1)
42
43 STATIC const uint16_t days_since_jan1[] = { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 };
44
timeutils_is_leap_year(mp_uint_t year)45 bool timeutils_is_leap_year(mp_uint_t year) {
46 return (year % 4 == 0 && year % 100 != 0) || year % 400 == 0;
47 }
48
49 // month is one based
timeutils_days_in_month(mp_uint_t year,mp_uint_t month)50 mp_uint_t timeutils_days_in_month(mp_uint_t year, mp_uint_t month) {
51 mp_uint_t mdays = days_since_jan1[month] - days_since_jan1[month - 1];
52 if (month == 2 && timeutils_is_leap_year(year)) {
53 mdays++;
54 }
55 return mdays;
56 }
57
58 // compute the day of the year, between 1 and 366
59 // month should be between 1 and 12, date should start at 1
timeutils_year_day(mp_uint_t year,mp_uint_t month,mp_uint_t date)60 mp_uint_t timeutils_year_day(mp_uint_t year, mp_uint_t month, mp_uint_t date) {
61 mp_uint_t yday = days_since_jan1[month - 1] + date;
62 if (month >= 3 && timeutils_is_leap_year(year)) {
63 yday += 1;
64 }
65 return yday;
66 }
67
timeutils_seconds_since_2000_to_struct_time(mp_uint_t t,timeutils_struct_time_t * tm)68 void timeutils_seconds_since_2000_to_struct_time(mp_uint_t t, timeutils_struct_time_t *tm) {
69 // The following algorithm was adapted from musl's __secs_to_tm and adapted
70 // for differences in MicroPython's timebase.
71
72 mp_int_t seconds = t - LEAPOCH;
73
74 mp_int_t days = seconds / 86400;
75 seconds %= 86400;
76 if (seconds < 0) {
77 seconds += 86400;
78 days -= 1;
79 }
80 tm->tm_hour = seconds / 3600;
81 tm->tm_min = seconds / 60 % 60;
82 tm->tm_sec = seconds % 60;
83
84 mp_int_t wday = (days + 2) % 7; // Mar 1, 2000 was a Wednesday (2)
85 if (wday < 0) {
86 wday += 7;
87 }
88 tm->tm_wday = wday;
89
90 mp_int_t qc_cycles = days / DAYS_PER_400Y;
91 days %= DAYS_PER_400Y;
92 if (days < 0) {
93 days += DAYS_PER_400Y;
94 qc_cycles--;
95 }
96 mp_int_t c_cycles = days / DAYS_PER_100Y;
97 if (c_cycles == 4) {
98 c_cycles--;
99 }
100 days -= (c_cycles * DAYS_PER_100Y);
101
102 mp_int_t q_cycles = days / DAYS_PER_4Y;
103 if (q_cycles == 25) {
104 q_cycles--;
105 }
106 days -= q_cycles * DAYS_PER_4Y;
107
108 mp_int_t years = days / 365;
109 if (years == 4) {
110 years--;
111 }
112 days -= (years * 365);
113
114 /* We will compute tm_yday at the very end
115 mp_int_t leap = !years && (q_cycles || !c_cycles);
116
117 tm->tm_yday = days + 31 + 28 + leap;
118 if (tm->tm_yday >= 365 + leap) {
119 tm->tm_yday -= 365 + leap;
120 }
121
122 tm->tm_yday++; // Make one based
123 */
124
125 tm->tm_year = 2000 + years + 4 * q_cycles + 100 * c_cycles + 400 * qc_cycles;
126
127 // Note: days_in_month[0] corresponds to March
128 STATIC const int8_t days_in_month[] = {31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 29};
129
130 mp_int_t month;
131 for (month = 0; days_in_month[month] <= days; month++) {
132 days -= days_in_month[month];
133 }
134
135 tm->tm_mon = month + 2;
136 if (tm->tm_mon >= 12) {
137 tm->tm_mon -= 12;
138 tm->tm_year++;
139 }
140 tm->tm_mday = days + 1; // Make one based
141 tm->tm_mon++; // Make one based
142
143 tm->tm_yday = timeutils_year_day(tm->tm_year, tm->tm_mon, tm->tm_mday);
144 }
145
146 // returns the number of seconds, as an integer, since 2000-01-01
timeutils_seconds_since_2000(mp_uint_t year,mp_uint_t month,mp_uint_t date,mp_uint_t hour,mp_uint_t minute,mp_uint_t second)147 mp_uint_t timeutils_seconds_since_2000(mp_uint_t year, mp_uint_t month,
148 mp_uint_t date, mp_uint_t hour, mp_uint_t minute, mp_uint_t second) {
149 return
150 second
151 + minute * 60
152 + hour * 3600
153 + (timeutils_year_day(year, month, date) - 1
154 + ((year - 2000 + 3) / 4) // add a day each 4 years starting with 2001
155 - ((year - 2000 + 99) / 100) // subtract a day each 100 years starting with 2001
156 + ((year - 2000 + 399) / 400) // add a day each 400 years starting with 2001
157 ) * 86400
158 + (year - 2000) * 31536000;
159 }
160
timeutils_mktime_2000(mp_uint_t year,mp_int_t month,mp_int_t mday,mp_int_t hours,mp_int_t minutes,mp_int_t seconds)161 mp_uint_t timeutils_mktime_2000(mp_uint_t year, mp_int_t month, mp_int_t mday,
162 mp_int_t hours, mp_int_t minutes, mp_int_t seconds) {
163
164 // Normalize the tuple. This allows things like:
165 //
166 // tm_tomorrow = list(time.localtime())
167 // tm_tomorrow[2] += 1 # Adds 1 to mday
168 // tomorrow = time.mktime(tm_tomorrow)
169 //
170 // And not have to worry about all the weird overflows.
171 //
172 // You can subtract dates/times this way as well.
173
174 minutes += seconds / 60;
175 if ((seconds = seconds % 60) < 0) {
176 seconds += 60;
177 minutes--;
178 }
179
180 hours += minutes / 60;
181 if ((minutes = minutes % 60) < 0) {
182 minutes += 60;
183 hours--;
184 }
185
186 mday += hours / 24;
187 if ((hours = hours % 24) < 0) {
188 hours += 24;
189 mday--;
190 }
191
192 month--; // make month zero based
193 year += month / 12;
194 if ((month = month % 12) < 0) {
195 month += 12;
196 year--;
197 }
198 month++; // back to one based
199
200 while (mday < 1) {
201 if (--month == 0) {
202 month = 12;
203 year--;
204 }
205 mday += timeutils_days_in_month(year, month);
206 }
207 while ((mp_uint_t)mday > timeutils_days_in_month(year, month)) {
208 mday -= timeutils_days_in_month(year, month);
209 if (++month == 13) {
210 month = 1;
211 year++;
212 }
213 }
214 return timeutils_seconds_since_2000(year, month, mday, hours, minutes, seconds);
215 }
216
217 // Calculate the weekday from the date.
218 // The result is zero based with 0 = Monday.
219 // by Michael Keith and Tom Craver, 1990.
timeutils_calc_weekday(int y,int m,int d)220 int timeutils_calc_weekday(int y, int m, int d) {
221 return ((d += m < 3 ? y-- : y - 2, 23 * m / 9 + d + 4 + y / 4 - y / 100 + y / 400) + 6) % 7;
222 }
223