clock_subr.c revision 1.16.30.1 1 /* $NetBSD: clock_subr.c,v 1.16.30.1 2014/11/12 18:50:55 snj Exp $ */
2
3 /*
4 * Copyright (c) 1988 University of Utah.
5 * Copyright (c) 1982, 1990, 1993
6 * The Regents of the University of California. All rights reserved.
7 *
8 * This code is derived from software contributed to Berkeley by
9 * the Systems Programming Group of the University of Utah Computer
10 * Science Department.
11 *
12 * Redistribution and use in source and binary forms, with or without
13 * modification, are permitted provided that the following conditions
14 * are met:
15 * 1. Redistributions of source code must retain the above copyright
16 * notice, this list of conditions and the following disclaimer.
17 * 2. Redistributions in binary form must reproduce the above copyright
18 * notice, this list of conditions and the following disclaimer in the
19 * documentation and/or other materials provided with the distribution.
20 * 3. Neither the name of the University nor the names of its contributors
21 * may be used to endorse or promote products derived from this software
22 * without specific prior written permission.
23 *
24 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
25 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
26 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
27 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
28 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
29 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
30 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
31 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
32 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
33 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
34 * SUCH DAMAGE.
35 *
36 * from: Utah $Hdr: clock.c 1.18 91/01/21$
37 *
38 * @(#)clock.c 8.2 (Berkeley) 1/12/94
39 */
40
41 /*
42 * Generic routines to convert between a POSIX date
43 * (seconds since 1/1/1970) and yr/mo/day/hr/min/sec
44 * Derived from arch/hp300/hp300/clock.c
45 */
46
47 #if HAVE_NBTOOL_CONFIG_H
48 #include "nbtool_config.h"
49 #endif /* HAVE_NBTOOL_CONFIG_H */
50
51 #ifdef _KERNEL
52 #include <sys/cdefs.h>
53 __KERNEL_RCSID(0, "$NetBSD: clock_subr.c,v 1.16.30.1 2014/11/12 18:50:55 snj Exp $");
54
55 #include <sys/param.h>
56 #include <sys/systm.h>
57 #include <sys/errno.h>
58 #else /* ! _KERNEL */
59 #include <string.h>
60 #include <time.h>
61 #include <errno.h>
62 #endif /* ! _KERNEL */
63
64 #include <dev/clock_subr.h>
65
66 static inline int leapyear(uint64_t year);
67 #define FEBRUARY 2
68 #define days_in_year(a) (leapyear(a) ? 366 : 365)
69 #define days_in_month(a) (month_days[(a) - 1])
70
71 /* for easier alignment:
72 * time from the epoch to 2000 (there were 7 leap years): */
73 #define DAYSTO2000 (365*30+7)
74
75 /* 4 year intervals include 1 leap year */
76 #define DAYS4YEARS (365*4+1)
77
78 /* 100 year intervals include 24 leap years */
79 #define DAYS100YEARS (365*100+24)
80
81 /* 400 year intervals include 97 leap years */
82 #define DAYS400YEARS (365*400+97)
83
84 static const int month_days[12] = {
85 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
86 };
87
88 /*
89 * This inline avoids some unnecessary modulo operations
90 * as compared with the usual macro:
91 * ( ((year % 4) == 0 &&
92 * (year % 100) != 0) ||
93 * ((year % 400) == 0) )
94 * It is otherwise equivalent.
95 */
96 static inline int
97 leapyear(uint64_t year)
98 {
99 int rv = 0;
100
101 if (year < 1969)
102 return EINVAL;
103
104 if ((year & 3) == 0) {
105 rv = 1;
106 if ((year % 100) == 0) {
107 rv = 0;
108 if ((year % 400) == 0)
109 rv = 1;
110 }
111 }
112 return rv;
113 }
114
115 time_t
116 clock_ymdhms_to_secs(struct clock_ymdhms *dt)
117 {
118 uint64_t secs, i, year, days;
119
120 year = dt->dt_year;
121
122 /*
123 * Compute days since start of time
124 * First from years, then from months.
125 */
126 if (year < POSIX_BASE_YEAR)
127 return -1;
128 days = 0;
129 if (leapyear(year) && dt->dt_mon > FEBRUARY)
130 days++;
131
132 if (year < 2000) {
133 /* simple way for early years */
134 for (i = POSIX_BASE_YEAR; i < year; i++)
135 days += days_in_year(i);
136 } else {
137 /* years are properly aligned */
138 days += DAYSTO2000;
139 year -= 2000;
140
141 i = year / 400;
142 days += i * DAYS400YEARS;
143 year -= i * 400;
144
145 i = year / 100;
146 days += i * DAYS100YEARS;
147 year -= i * 100;
148
149 i = year / 4;
150 days += i * DAYS4YEARS;
151 year -= i * 4;
152
153 for (i = dt->dt_year-year; i < dt->dt_year; i++)
154 days += days_in_year(i);
155 }
156
157
158 /* Months */
159 for (i = 1; i < dt->dt_mon; i++)
160 days += days_in_month(i);
161 days += (dt->dt_day - 1);
162
163 /* Add hours, minutes, seconds. */
164 secs = (((uint64_t)days
165 * 24 + dt->dt_hour)
166 * 60 + dt->dt_min)
167 * 60 + dt->dt_sec;
168
169 if ((time_t)secs < 0 || secs > __type_max(time_t))
170 return -1;
171 return secs;
172 }
173
174 int
175 clock_secs_to_ymdhms(time_t secs, struct clock_ymdhms *dt)
176 {
177 int leap;
178 uint64_t i;
179 time_t days;
180 time_t rsec; /* remainder seconds */
181
182 if (secs < 0)
183 return EINVAL;
184
185 days = secs / SECDAY;
186 rsec = secs % SECDAY;
187
188 /* Day of week (Note: 1/1/1970 was a Thursday) */
189 dt->dt_wday = (days + 4) % 7;
190
191 if (days >= DAYSTO2000) {
192 days -= DAYSTO2000;
193 dt->dt_year = 2000;
194
195 i = days / DAYS400YEARS;
196 days -= i*DAYS400YEARS;
197 dt->dt_year += i*400;
198
199 i = days / DAYS100YEARS;
200 days -= i*DAYS100YEARS;
201 dt->dt_year += i*100;
202
203 i = days / DAYS4YEARS;
204 days -= i*DAYS4YEARS;
205 dt->dt_year += i*4;
206
207 for (i = dt->dt_year; days >= days_in_year(i); i++)
208 days -= days_in_year(i);
209 dt->dt_year = i;
210 } else {
211 /* Subtract out whole years, counting them in i. */
212 for (i = POSIX_BASE_YEAR; days >= days_in_year(i); i++)
213 days -= days_in_year(i);
214 dt->dt_year = i;
215 }
216
217 /* Subtract out whole months, counting them in i. */
218 for (leap = 0, i = 1; days >= days_in_month(i)+leap; i++) {
219 days -= days_in_month(i)+leap;
220 if (i == 1 && leapyear(dt->dt_year))
221 leap = 1;
222 else
223 leap = 0;
224 }
225 dt->dt_mon = i;
226
227 /* Days are what is left over (+1) from all that. */
228 dt->dt_day = days + 1;
229
230 /* Hours, minutes, seconds are easy */
231 dt->dt_hour = rsec / 3600;
232 rsec = rsec % 3600;
233 dt->dt_min = rsec / 60;
234 rsec = rsec % 60;
235 dt->dt_sec = rsec;
236
237 return 0;
238 }
239