Contributors: 3
Author Tokens Token Proportion Commits Commit Proportion
Zhao Lei 480 96.77% 1 33.33%
Deepa Dinamani 15 3.02% 1 33.33%
Thomas Gleixner 1 0.20% 1 33.33%
Total 496 3


// SPDX-License-Identifier: LGPL-2.0+
/*
 * 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)

/**
 * time64_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 time64_to_tm(time64_t totalsecs, int offset, struct tm *result)
{
	long days, rem, y;
	int remainder;
	const unsigned short *ip;

	days = div_s64_rem(totalsecs, SECS_PER_DAY, &remainder);
	rem = remainder;
	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(time64_to_tm);