linux/kernel/time/timeconv.c
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   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 */
  37static 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 */
  43static 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 */
  49static 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). */
  59static 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 * time_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 */
  77void time_to_tm(time_t totalsecs, int offset, struct tm *result)
  78{
  79        long days, rem, y;
  80        const unsigned short *ip;
  81
  82        days = totalsecs / SECS_PER_DAY;
  83        rem = totalsecs % SECS_PER_DAY;
  84        rem += offset;
  85        while (rem < 0) {
  86                rem += SECS_PER_DAY;
  87                --days;
  88        }
  89        while (rem >= SECS_PER_DAY) {
  90                rem -= SECS_PER_DAY;
  91                ++days;
  92        }
  93
  94        result->tm_hour = rem / SECS_PER_HOUR;
  95        rem %= SECS_PER_HOUR;
  96        result->tm_min = rem / 60;
  97        result->tm_sec = rem % 60;
  98
  99        /* January 1, 1970 was a Thursday. */
 100        result->tm_wday = (4 + days) % 7;
 101        if (result->tm_wday < 0)
 102                result->tm_wday += 7;
 103
 104        y = 1970;
 105
 106        while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
 107                /* Guess a corrected year, assuming 365 days per year. */
 108                long yg = y + math_div(days, 365);
 109
 110                /* Adjust DAYS and Y to match the guessed year. */
 111                days -= (yg - y) * 365 + leaps_between(y, yg);
 112                y = yg;
 113        }
 114
 115        result->tm_year = y - 1900;
 116
 117        result->tm_yday = days;
 118
 119        ip = __mon_yday[__isleap(y)];
 120        for (y = 11; days < ip[y]; y--)
 121                continue;
 122        days -= ip[y];
 123
 124        result->tm_mon = y;
 125        result->tm_mday = days + 1;
 126}
 127EXPORT_SYMBOL(time_to_tm);
 128
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