https://github.com/torvalds/linux
Tip revision: 8f0d8163b50e01f398b14bcd4dc039ac5ab18d64 authored by Linus Torvalds on 28 October 2012, 19:24:48 UTC
Linux 3.7-rc3
Linux 3.7-rc3
Tip revision: 8f0d816
timeconv.c
/*
* Copyright (C) 1993, 1994, 1995, 1996, 1997 Free Software Foundation, Inc.
* This file is part of the GNU C Library.
* Contributed by Paul Eggert (eggert@twinsun.com).
*
* The GNU C Library is free software; you can redistribute it and/or
* modify it under the terms of the GNU Library General Public License as
* published by the Free Software Foundation; either version 2 of the
* License, or (at your option) any later version.
*
* The GNU C Library is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
* Library General Public License for more details.
*
* You should have received a copy of the GNU Library General Public
* License along with the GNU C Library; see the file COPYING.LIB. If not,
* write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
* Boston, MA 02111-1307, USA.
*/
/*
* Converts the calendar time to broken-down time representation
* Based on code from glibc-2.6
*
* 2009-7-14:
* Moved from glibc-2.6 to kernel by Zhaolei<zhaolei@cn.fujitsu.com>
*/
#include <linux/time.h>
#include <linux/module.h>
/*
* Nonzero if YEAR is a leap year (every 4 years,
* except every 100th isn't, and every 400th is).
*/
static int __isleap(long year)
{
return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
}
/* do a mathdiv for long type */
static long math_div(long a, long b)
{
return a / b - (a % b < 0);
}
/* How many leap years between y1 and y2, y1 must less or equal to y2 */
static long leaps_between(long y1, long y2)
{
long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
+ math_div(y1 - 1, 400);
long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
+ math_div(y2 - 1, 400);
return leaps2 - leaps1;
}
/* How many days come before each month (0-12). */
static const unsigned short __mon_yday[2][13] = {
/* Normal years. */
{0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
/* Leap years. */
{0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
};
#define SECS_PER_HOUR (60 * 60)
#define SECS_PER_DAY (SECS_PER_HOUR * 24)
/**
* time_to_tm - converts the calendar time to local broken-down time
*
* @totalsecs the number of seconds elapsed since 00:00:00 on January 1, 1970,
* Coordinated Universal Time (UTC).
* @offset offset seconds adding to totalsecs.
* @result pointer to struct tm variable to receive broken-down time
*/
void time_to_tm(time_t totalsecs, int offset, struct tm *result)
{
long days, rem, y;
const unsigned short *ip;
days = totalsecs / SECS_PER_DAY;
rem = totalsecs % SECS_PER_DAY;
rem += offset;
while (rem < 0) {
rem += SECS_PER_DAY;
--days;
}
while (rem >= SECS_PER_DAY) {
rem -= SECS_PER_DAY;
++days;
}
result->tm_hour = rem / SECS_PER_HOUR;
rem %= SECS_PER_HOUR;
result->tm_min = rem / 60;
result->tm_sec = rem % 60;
/* January 1, 1970 was a Thursday. */
result->tm_wday = (4 + days) % 7;
if (result->tm_wday < 0)
result->tm_wday += 7;
y = 1970;
while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
/* Guess a corrected year, assuming 365 days per year. */
long yg = y + math_div(days, 365);
/* Adjust DAYS and Y to match the guessed year. */
days -= (yg - y) * 365 + leaps_between(y, yg);
y = yg;
}
result->tm_year = y - 1900;
result->tm_yday = days;
ip = __mon_yday[__isleap(y)];
for (y = 11; days < ip[y]; y--)
continue;
days -= ip[y];
result->tm_mon = y;
result->tm_mday = days + 1;
}
EXPORT_SYMBOL(time_to_tm);