Julian date routines needed

[Neil Woods]

In article <1753 at manta.NOSC.MIL> grantk at manta.NOSC.MIL (Kelly J. Grant) writes:
>Howdy networld
>
>About 100,000 years ago I had some Julian date routines (in dBASE) to
>convert YY/MM/DD into a format suitable for math, and also to convert
>back to normal dates.  We called these JULTOCAL and CALTOJUL.  I now
>have a need for these algorithms for a UNIX program.  Would any of 
>you kind souls have these routines lying around in your book of tricks
>[right next to your hanoi.c file maybe :-)]

The following should do the trick:

------------(Cut Here)------------

#include <stdio.h>

/* returns Julian day number --given a string of the form xx/yy/zz.
** To calculate the day of the week use the return value % 7.
*/

long 
julian_day (date_str)
  char *date_str;
{
  long jdn;
  int month, day, year;

  sscanf (date_str, "%d/%d/%d", &month, &day, &year);

  if (year < 100)
    year += 1900;          /* assume 20th century */
  if (year < 1000)
    year += 2000;          /* assume 21st */
  jdn = (long) year * 367 + month * 275 / 9
  - (year + (month > 2)) * 7 / 4
  - ((year - (month < 3)) / 100 + 1) * 3 / 4 + day + 1721029L;

  return (jdn);
}

/* converts from a given Julian day number to a formatted string of the form
** xx/yy/zz and places it in date_string. Returns the century (e.g. 19).
*/

int 
julian_date (jdn, date_string)
  long jdn;
  char *date_string;
{
  long year, month, day, temp_var;
  int century;

  temp_var = jdn - 1721119L;
  year = (4 * temp_var - 1) / 146097L;
  temp_var = 4 * temp_var - 1 - 146097L * year;
  day = temp_var / 4;
  temp_var = (4 * day + 3) / 1461;
  day = 4 * day + 3 - 1461 * temp_var;
  day = (day + 4) / 4;
  month = (5 * day - 3) / 153;
  day = 5 * day - 3 - 153 * month;
  day = (day + 5) / 5;
  year = 100 * year + temp_var;
  if (month < 10)
    month += 3;
  else
    {
      month -= 9;
      year++;
    }

  century = (int)year / 100;
  year = year - (century * 100);

  sprintf (date_string, "%02ld/%02ld/%02ld", month, day, year);

  return ((int) century);
}

/* Return the difference between two dates in days as a signed long */

long 
diff_date (date1, date2)
  char *date1, *date2;
{
  return (julian_day (date2) - julian_day (date1));
}

int
main (argc, argv)
  int argc;
  char *argv[];
{
  if (argc != 3)
    {
      fprintf (stderr, "Usage: diffdate first_date second_date\n");
      fprintf (stderr, "Where each date is of the form mm/dd/yy\n");
      return 1;
    }
   printf ("Difference in days = %ld\n", diff_date (argv[1], argv[2]));
   return 0;
}


>Muchos gracias in advance
>
>Kelly

Your Welcome.

	Neil.

-- 
Neil Woods.                 | JANET: ex1neil at uk.ac.swan.pyr
University College Swansea, | UUCP : ...!mcvax!ukc!swan.pyr!ex1neil
Wales, UK.                  | ARPA : ex1neil%pyr.swan.ac.uk at nfsnet-relay.ac.uk