kernel/time/timeconv.c

   1 /*
   2  * Copyright (C) 1993, 1994, 1995, 1996, 1997 Free Software Foundation, Inc.
   3  * This file is part of the GNU C Library.
   4  * Contributed by Paul Eggert (eggert@twinsun.com).
   5  *
   6  * The GNU C Library is free software; you can redistribute it and/or
   7  * modify it under the terms of the GNU Library General Public License as
   8  * published by the Free Software Foundation; either version 2 of the
   9  * License, or (at your option) any later version.
  10  *
  11  * The GNU C Library is distributed in the hope that it will be useful,
  12  * but WITHOUT ANY WARRANTY; without even the implied warranty of
  13  * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  14  * Library General Public License for more details.
  15  *
  16  * You should have received a copy of the GNU Library General Public
  17  * License along with the GNU C Library; see the file COPYING.LIB.  If not,
  18  * write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
  19  * Boston, MA 02111-1307, USA.
  20  */
  21
  22 /*
  23  * Converts the calendar time to broken-down time representation
  24  * Based on code from glibc-2.6
  25  *
  26  * 2009-7-14:
  27  *   Moved from glibc-2.6 to kernel by Zhaolei<zhaolei@cn.fujitsu.com>
  28  */
  29
  30 #include <linux/time.h>
  31 #include <linux/module.h>
  32
  33 /*
  34  * Nonzero if YEAR is a leap year (every 4 years,
  35  * except every 100th isn't, and every 400th is).
  36  */
  37 static int __isleap(long year)
  38 {
  39         return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
  40 }
  41
  42 /* do a mathdiv for long type */
  43 static long math_div(long a, long b)
  44 {
  45         return a / b - (a % b < 0);
  46 }
  47
  48 /* How many leap years between y1 and y2, y1 must less or equal to y2 */
  49 static long leaps_between(long y1, long y2)
  50 {
  51         long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
  52                 + math_div(y1 - 1, 400);
  53         long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
  54                 + math_div(y2 - 1, 400);
  55         return leaps2 - leaps1;
  56 }
  57
  58 /* How many days come before each month (0-12). */
  59 static const unsigned short __mon_yday[2][13] = {
  60         /* Normal years. */
  61         {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
  62         /* Leap years. */
  63         {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
  64 };
  65
  66 #define SECS_PER_HOUR   (60 * 60)
  67 #define SECS_PER_DAY    (SECS_PER_HOUR * 24)
  68
  69 /**
  70  * time64_to_tm - converts the calendar time to local broken-down time
  71  *
  72  * @totalsecs   the number of seconds elapsed since 00:00:00 on January 1, 1970,
  73  *              Coordinated Universal Time (UTC).
  74  * @offset      offset seconds adding to totalsecs.
  75  * @result      pointer to struct tm variable to receive broken-down time
  76  */
  77 void time64_to_tm(time64_t totalsecs, int offset, struct tm *result)
  78 {
  79         long days, rem, y;
  80         int remainder;
  81         const unsigned short *ip;
  82
  83         days = div_s64_rem(totalsecs, SECS_PER_DAY, &remainder);
  84         rem = remainder;
  85         rem += offset;
  86         while (rem < 0) {
  87                 rem += SECS_PER_DAY;
  88                 --days;
  89         }
  90         while (rem >= SECS_PER_DAY) {
  91                 rem -= SECS_PER_DAY;
  92                 ++days;
  93         }
  94
  95         result->tm_hour = rem / SECS_PER_HOUR;
  96         rem %= SECS_PER_HOUR;
  97         result->tm_min = rem / 60;
  98         result->tm_sec = rem % 60;
  99
 100         /* January 1, 1970 was a Thursday. */
 101         result->tm_wday = (4 + days) % 7;
 102         if (result->tm_wday < 0)
 103                 result->tm_wday += 7;
 104
 105         y = 1970;
 106
 107         while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
 108                 /* Guess a corrected year, assuming 365 days per year. */
 109                 long yg = y + math_div(days, 365);
 110
 111                 /* Adjust DAYS and Y to match the guessed year. */
 112                 days -= (yg - y) * 365 + leaps_between(y, yg);
 113                 y = yg;
 114         }
 115
 116         result->tm_year = y - 1900;
 117
 118         result->tm_yday = days;
 119
 120         ip = __mon_yday[__isleap(y)];
 121         for (y = 11; days < ip[y]; y--)
 122                 continue;
 123         days -= ip[y];
 124
 125         result->tm_mon = y;
 126         result->tm_mday = days + 1;
 127 }
 128 EXPORT_SYMBOL(time64_to_tm);