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