timeconv.c 3.6 KB

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129
  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 (eggert@twinsun.com).
  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. * Converts the calendar time to broken-down time representation
  24. * Based on code from glibc-2.6
  25. *
  26. * 2009-7-14:
  27. * Moved from glibc-2.6 to kernel by Zhaolei<zhaolei@cn.fujitsu.com>
  28. */
  29. #include <linux/time.h>
  30. #include <linux/module.h>
  31. /*
  32. * Nonzero if YEAR is a leap year (every 4 years,
  33. * except every 100th isn't, and every 400th is).
  34. */
  35. static int __isleap(long year)
  36. {
  37. return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
  38. }
  39. /* do a mathdiv for long type */
  40. static long math_div(long a, long b)
  41. {
  42. return a / b - (a % b < 0);
  43. }
  44. /* How many leap years between y1 and y2, y1 must less or equal to y2 */
  45. static long leaps_between(long y1, long y2)
  46. {
  47. long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
  48. + math_div(y1 - 1, 400);
  49. long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
  50. + math_div(y2 - 1, 400);
  51. return leaps2 - leaps1;
  52. }
  53. /* How many days come before each month (0-12). */
  54. static const unsigned short __mon_yday[2][13] = {
  55. /* Normal years. */
  56. {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
  57. /* Leap years. */
  58. {0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
  59. };
  60. #define SECS_PER_HOUR (60 * 60)
  61. #define SECS_PER_DAY (SECS_PER_HOUR * 24)
  62. /**
  63. * time64_to_tm - converts the calendar time to local broken-down time
  64. *
  65. * @totalsecs the number of seconds elapsed since 00:00:00 on January 1, 1970,
  66. * Coordinated Universal Time (UTC).
  67. * @offset offset seconds adding to totalsecs.
  68. * @result pointer to struct tm variable to receive broken-down time
  69. */
  70. void time64_to_tm(time64_t totalsecs, int offset, struct tm *result)
  71. {
  72. long days, rem, y;
  73. int remainder;
  74. const unsigned short *ip;
  75. days = div_s64_rem(totalsecs, SECS_PER_DAY, &remainder);
  76. rem = remainder;
  77. rem += offset;
  78. while (rem < 0) {
  79. rem += SECS_PER_DAY;
  80. --days;
  81. }
  82. while (rem >= SECS_PER_DAY) {
  83. rem -= SECS_PER_DAY;
  84. ++days;
  85. }
  86. result->tm_hour = rem / SECS_PER_HOUR;
  87. rem %= SECS_PER_HOUR;
  88. result->tm_min = rem / 60;
  89. result->tm_sec = rem % 60;
  90. /* January 1, 1970 was a Thursday. */
  91. result->tm_wday = (4 + days) % 7;
  92. if (result->tm_wday < 0)
  93. result->tm_wday += 7;
  94. y = 1970;
  95. while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
  96. /* Guess a corrected year, assuming 365 days per year. */
  97. long yg = y + math_div(days, 365);
  98. /* Adjust DAYS and Y to match the guessed year. */
  99. days -= (yg - y) * 365 + leaps_between(y, yg);
  100. y = yg;
  101. }
  102. result->tm_year = y - 1900;
  103. result->tm_yday = days;
  104. ip = __mon_yday[__isleap(y)];
  105. for (y = 11; days < ip[y]; y--)
  106. continue;
  107. days -= ip[y];
  108. result->tm_mon = y;
  109. result->tm_mday = days + 1;
  110. }
  111. EXPORT_SYMBOL(time64_to_tm);