cregit-Linux how code gets into the kernel

Release 4.15 kernel/time/timeconv.c

Directory: kernel/time
/*
 * 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); }

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Zhao Lei36100.00%1100.00%
Total36100.00%1100.00%

/* do a mathdiv for long type */
static long math_div(long a, long b) { return a / b - (a % b < 0); }

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Zhao Lei25100.00%1100.00%
Total25100.00%1100.00%

/* 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; }

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Zhao Lei77100.00%1100.00%
Total77100.00%1100.00%

/* 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; }

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Zhao Lei24694.98%150.00%
Deepa Dinamani135.02%150.00%
Total259100.00%2100.00%

EXPORT_SYMBOL(time64_to_tm);

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Zhao Lei48096.97%150.00%
Deepa Dinamani153.03%150.00%
Total495100.00%2100.00%
Directory: kernel/time
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