kernel/time/timeconv.c

   1 // SPDX-License-Identifier: LGPL-2.0+
   2 /*
   3  * Copyright (C) 1993, 1994, 1995, 1996, 1997 Free Software Foundation, Inc.
   4  * This file is part of the GNU C Library.
   5  * Contributed by Paul Eggert (eggert@twinsun.com).
   6  *
   7  * The GNU C Library is free software; you can redistribute it and/or
   8  * modify it under the terms of the GNU Library General Public License as
   9  * published by the Free Software Foundation; either version 2 of the
  10  * License, or (at your option) any later version.
  11  *
  12  * The GNU C Library is distributed in the hope that it will be useful,
  13  * but WITHOUT ANY WARRANTY; without even the implied warranty of
  14  * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  15  * Library General Public License for more details.
  16  *
  17  * You should have received a copy of the GNU Library General Public
  18  * License along with the GNU C Library; see the file COPYING.LIB.  If not,
  19  * write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
  20  * Boston, MA 02111-1307, USA.
  21  */
  22
  23 /*
  24  * Converts the calendar time to broken-down time representation
  25  * Based on code from glibc-2.6
  26  *
  27  * 2009-7-14:
  28  *   Moved from glibc-2.6 to kernel by Zhaolei<zhaolei@cn.fujitsu.com>
  29  */
  30
  31 #include <linux/time.h>
  32 #include <linux/module.h>
  33
  34 /*
  35  * Nonzero if YEAR is a leap year (every 4 years,
  36  * except every 100th isn't, and every 400th is).
  37  */
  38 static int __isleap(long year)
  39 {
  40         return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
  41 }
  42
  43 /* do a mathdiv for long type */
  44 static long math_div(long a, long b)
  45 {
  46         return a / b - (a % b < 0);
  47 }
  48
  49 /* How many leap years between y1 and y2, y1 must less or equal to y2 */
  50 static long leaps_between(long y1, long y2)
  51 {
  52         long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
  53                 + math_div(y1 - 1, 400);
  54         long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
  55                 + math_div(y2 - 1, 400);
  56         return leaps2 - leaps1;
  57 }
  58
  59 /* How many days come before each month (0-12). */
  60 static const unsigned short __mon_yday[2][13] = {
  61         /* Normal years. */
  62         {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
  63         /* Leap years. */
  64         {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
  65 };
  66
  67 #define SECS_PER_HOUR   (60 * 60)
  68 #define SECS_PER_DAY    (SECS_PER_HOUR * 24)
  69
  70 /**
  71  * time64_to_tm - converts the calendar time to local broken-down time
  72  *
  73  * @totalsecs:  the number of seconds elapsed since 00:00:00 on January 1, 1970,
  74  *              Coordinated Universal Time (UTC).
  75  * @offset:     offset seconds adding to totalsecs.
  76  * @result:     pointer to struct tm variable to receive broken-down time
  77  */
  78 void time64_to_tm(time64_t totalsecs, int offset, struct tm *result)
  79 {
  80         long days, rem, y;
  81         int remainder;
  82         const unsigned short *ip;
  83
  84         days = div_s64_rem(totalsecs, SECS_PER_DAY, &remainder);
  85         rem = remainder;
  86         rem += offset;
  87         while (rem < 0) {
  88                 rem += SECS_PER_DAY;
  89                 --days;
  90         }
  91         while (rem >= SECS_PER_DAY) {
  92                 rem -= SECS_PER_DAY;
  93                 ++days;
  94         }
  95
  96         result->tm_hour = rem / SECS_PER_HOUR;
  97         rem %= SECS_PER_HOUR;
  98         result->tm_min = rem / 60;
  99         result->tm_sec = rem % 60;
 100
 101         /* January 1, 1970 was a Thursday. */
 102         result->tm_wday = (4 + days) % 7;
 103         if (result->tm_wday < 0)
 104                 result->tm_wday += 7;
 105
 106         y = 1970;
 107
 108         while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
 109                 /* Guess a corrected year, assuming 365 days per year. */
 110                 long yg = y + math_div(days, 365);
 111
 112                 /* Adjust DAYS and Y to match the guessed year. */
 113                 days -= (yg - y) * 365 + leaps_between(y, yg);
 114                 y = yg;
 115         }
 116
 117         result->tm_year = y - 1900;
 118
 119         result->tm_yday = days;
 120
 121         ip = __mon_yday[__isleap(y)];
 122         for (y = 11; days < ip[y]; y--)
 123                 continue;
 124         days -= ip[y];
 125
 126         result->tm_mon = y;
 127         result->tm_mday = days + 1;
 128 }
 129 EXPORT_SYMBOL(time64_to_tm);