offtime.c

Linux下头文件time.h的实现源码

/* Copyright (C) 1991, 1993, 1997, 1998, 2002 Free Software Foundation, Inc.
   This file is part of the GNU C Library.

   The GNU C Library is free software; you can redistribute it and/or
   modify it under the terms of the GNU Lesser General Public
   License as published by the Free Software Foundation; either
   version 2.1 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
   Lesser General Public License for more details.

   You should have received a copy of the GNU Lesser General Public
   License along with the GNU C Library; if not, write to the Free
   Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
   02111-1307 USA.  */

#include <errno.h>
#include <time.h>

#define	SECS_PER_HOUR	(60 * 60)
#define	SECS_PER_DAY	(SECS_PER_HOUR * 24)

/* Compute the `struct tm' representation of *T,
   offset OFFSET seconds east of UTC,
   and store year, yday, mon, mday, wday, hour, min, sec into *TP.
   Return nonzero if successful.  */
int
__offtime (t, offset, tp)
     const time_t *t;
     long int offset;
     struct tm *tp;
{
  long int days, rem, y;
  const unsigned short int *ip;

  days = *t / SECS_PER_DAY;
  rem = *t % SECS_PER_DAY;
  rem += offset;
  while (rem < 0)
    {
      rem += SECS_PER_DAY;
      --days;
    }
  while (rem >= SECS_PER_DAY)
    {
      rem -= SECS_PER_DAY;
      ++days;
    }
  tp->tm_hour = rem / SECS_PER_HOUR;
  rem %= SECS_PER_HOUR;
  tp->tm_min = rem / 60;
  tp->tm_sec = rem % 60;
  /* January 1, 1970 was a Thursday.  */
  tp->tm_wday = (4 + days) % 7;
  if (tp->tm_wday < 0)
    tp->tm_wday += 7;
  y = 1970;

#define DIV(a, b) ((a) / (b) - ((a) % (b) < 0))
#define LEAPS_THRU_END_OF(y) (DIV (y, 4) - DIV (y, 100) + DIV (y, 400))

  while (days < 0 || days >= (__isleap (y) ? 366 : 365))
    {
      /* Guess a corrected year, assuming 365 days per year.  */
      long int yg = y + days / 365 - (days % 365 < 0);

      /* Adjust DAYS and Y to match the guessed year.  */
      days -= ((yg - y) * 365
	       + LEAPS_THRU_END_OF (yg - 1)
	       - LEAPS_THRU_END_OF (y - 1));
      y = yg;
    }
  tp->tm_year = y - 1900;
  if (tp->tm_year != y - 1900)
    {
      /* The year cannot be represented due to overflow.  */
      __set_errno (EOVERFLOW);
      return 0;
    }
  tp->tm_yday = days;
  ip = __mon_yday[__isleap(y)];
  for (y = 11; days < (long int) ip[y]; --y)
    continue;
  days -= ip[y];
  tp->tm_mon = y;
  tp->tm_mday = days + 1;
  return 1;
}