1 /* $NetBSD: clock_subr.c,v 1.22 2014/09/07 11:50:23 martin 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 1.22 2014/09/07 11:50:23 martin 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