projects/linux·blob v5.7 · refs

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1// SPDX-License-Identifier: LGPL-2.0+
2/*
3 * Copyright (C) 1993, 1994, 1995, 1996, 1997 Free Software Foundation, Inc.
4 * This file is part of the GNU C Library.
5 * Contributed by Paul Eggert ([email protected]).
6 *
7 * The GNU C Library is free software; you can redistribute it and/or
8 * modify it under the terms of the GNU Library General Public License as
9 * published by the Free Software Foundation; either version 2 of the
10 * License, or (at your option) any later version.
11 *
12 * The GNU C Library is distributed in the hope that it will be useful,
13 * but WITHOUT ANY WARRANTY; without even the implied warranty of
14 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
15 * Library General Public License for more details.
16 *
17 * You should have received a copy of the GNU Library General Public
18 * License along with the GNU C Library; see the file COPYING.LIB. If not,
19 * write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
20 * Boston, MA 02111-1307, USA.
21 */
22
23/*
24 * Converts the calendar time to broken-down time representation
25 * Based on code from glibc-2.6
26 *
27 * 2009-7-14:
28 * Moved from glibc-2.6 to kernel by Zhaolei<[email protected]>
29 */
30
31#include <linux/time.h>
32#include <linux/module.h>
33
34/*
35 * Nonzero if YEAR is a leap year (every 4 years,
36 * except every 100th isn't, and every 400th is).
37 */
38static int __isleap(long year)
39{
40 return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
41}
42
43/* do a mathdiv for long type */
44static long math_div(long a, long b)
45{
46 return a / b - (a % b < 0);
47}
48
49/* How many leap years between y1 and y2, y1 must less or equal to y2 */
50static long leaps_between(long y1, long y2)
51{
52 long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
53 + math_div(y1 - 1, 400);
54 long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
55 + math_div(y2 - 1, 400);
56 return leaps2 - leaps1;
57}
58
59/* How many days come before each month (0-12). */
60static const unsigned short __mon_yday[2][13] = {
61 /* Normal years. */
62 {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
63 /* Leap years. */
64 {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
65};
66
67#define SECS_PER_HOUR (60 * 60)
68#define SECS_PER_DAY (SECS_PER_HOUR * 24)
69
70/**
71 * time64_to_tm - converts the calendar time to local broken-down time
72 *
73 * @totalsecs the number of seconds elapsed since 00:00:00 on January 1, 1970,
74 * Coordinated Universal Time (UTC).
75 * @offset offset seconds adding to totalsecs.
76 * @result pointer to struct tm variable to receive broken-down time
77 */
78void time64_to_tm(time64_t totalsecs, int offset, struct tm *result)
79{
80 long days, rem, y;
81 int remainder;
82 const unsigned short *ip;
83
84 days = div_s64_rem(totalsecs, SECS_PER_DAY, &remainder);
85 rem = remainder;
86 rem += offset;
87 while (rem < 0) {
88 rem += SECS_PER_DAY;
89 --days;
90 }
91 while (rem >= SECS_PER_DAY) {
92 rem -= SECS_PER_DAY;
93 ++days;
94 }
95
96 result->tm_hour = rem / SECS_PER_HOUR;
97 rem %= SECS_PER_HOUR;
98 result->tm_min = rem / 60;
99 result->tm_sec = rem % 60;
100
101 /* January 1, 1970 was a Thursday. */
102 result->tm_wday = (4 + days) % 7;
103 if (result->tm_wday < 0)
104 result->tm_wday += 7;
105
106 y = 1970;
107
108 while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
109 /* Guess a corrected year, assuming 365 days per year. */
110 long yg = y + math_div(days, 365);
111
112 /* Adjust DAYS and Y to match the guessed year. */
113 days -= (yg - y) * 365 + leaps_between(y, yg);
114 y = yg;
115 }
116
117 result->tm_year = y - 1900;
118
119 result->tm_yday = days;
120
121 ip = __mon_yday[__isleap(y)];
122 for (y = 11; days < ip[y]; y--)
123 continue;
124 days -= ip[y];
125
126 result->tm_mon = y;
127 result->tm_mday = days + 1;
128}
129EXPORT_SYMBOL(time64_to_tm);