Пример #1
0
/****************************************************************************
   function: test_2
 parameters: none
    returns: if successful
	       TRUE
	     else
	       FALSE
    purpose: The following code will test the days_to_date() function.
****************************************************************************/
char test_2(FILE *output)
  {
  #define DAYS_TEST_DATES 5
  char status = PASS,
       b_date[7],
       e_date[7],
       *answer_table[DAYS_TEST_DATES];
  long days,
       i;

  /* build the correct answer table */
  answer_table[0] = "910101";
  answer_table[1] = "910131";
  answer_table[2] = "910302";
  answer_table[3] = "910401";
  answer_table[4] = "910501";
  strcpy(b_date, "910101");
  for (days = 0, i = 0; i < DAYS_TEST_DATES; days += 30, i++)
    {
    days_to_date(b_date, days, e_date);
    if (strcmp(e_date, answer_table[i]) != 0)
      {
      fprintf(output,
	      "ERROR: beginning [%s] + days [%ld] = ending [%s] [%s]\n",
	      b_date, days, e_date, answer_table[i]);
      status = FALSE;
      }
    }
  return(status);
  }
Пример #2
0
int main (int argc, char *argv [])
{
    long date, time;

    date = date_now ();
    time = time_now ();
    printf ("Date=%ld time=%ld\n", date, time);

    date = days_to_date  (date_to_days (date));
    time = csecs_to_time (time_to_csecs (time));
    printf ("Date=%ld time=%ld\n", date, time);

    date = date_now ();
    time = time_now ();
    date = days_to_date  (date_to_days (date));
    time = csecs_to_time (time_to_csecs (time));

    future_date (&date, &time, 0, INTERVAL_HOUR);
    printf ("Date in one hour = %ld, %ld\n", date, time);

    date = date_now ();
    time = time_now ();
    future_date (&date, &time, 0, INTERVAL_DAY);
    printf ("Date in one day  = %ld, %ld\n", date, time);

    date = date_now ();
    time = time_now ();
    future_date (&date, &time, 1, 0);
    printf ("Date in one day  = %ld, %ld\n", date, time);

    date = date_now ();
    time = time_now ();
    future_date (&date, &time, 7, 0);
    printf ("Date in one week = %ld, %ld\n", date, time);

    date = date_now ();
    time = time_now ();
    future_date (&date, &time, 365, 0);
    printf ("Date in one year = %ld, %ld\n", date, time);

    return (EXIT_SUCCESS);
}
Пример #3
0
int main()
{
    int days, d, m, y;

    while (true) {
        scanf("%d%d%d%d", &days, &d, &m, &y);
        if (y == 0) break;

        int x = date_to_days(d, m, y);
        days_to_date(x + days, d, m, y);

        printf("%d %d %d\n", d, m, y);
    }

    return 0;
}