Lines Matching full:year
91 * Note that 1 year has < 2^25 seconds. So an PRInt32 is large enough.
134 gregorian_date.year = exploded->tm_year;
172 // the most future date possible (year 2038).
176 // the most past date possible (year 1901).
193 * till the start of the given year Y. At the start of the year 4
195 * the year 5 A.D. this count is 1. The number of years divisible by
200 * start of the given year Y. The number of days at the start of the year
201 * 1 is 0 while the number of days at the start of the year 2 is 365
215 * The following array contains the day of year for the last day of
238 * Returns 1 if the year is a leap year, 0 otherwise.
243 static int IsLeapYear(PRInt16 year)
245 if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
372 /* Normalize month and year before mday */
382 /* Now that month and year are in proper range, normalize mday */
529 PRInt32 year = -1;
927 year = n1;
928 if (year < 70)
929 year += 2000;
930 else if (year < 100)
931 year += 1900;
953 year = n3;
961 year = n3;
969 else if ((end - rest) == 5) /* five digits is a year */
970 year = (year < 0
976 : year);
977 else if ((end - rest) == 4) /* four digits is a year */
978 year = (year < 0
983 : year);
984 else if ((end - rest) == 2) /* two digits - date or year */
991 Otherwise, if we have a date and not a year, assume this is the
992 year. If it is less than 70, then assume it refers to the 21st
994 century. Otherwise, assume it refers to an unambiguous year.
1000 else if (year < 0)
1003 year = 2000 + n;
1005 year = 1900 + n;
1007 year = n;
1039 parsed a year (e.g., the second "-" in "30-AUG-1966"), or if
1041 if (*rest == '-' && ((rest > string && isalpha(rest[-1]) && year < 0)
1075 /* If we didn't find a year, month, or day-of-the-month, we can't
1079 if (month == TT_UNKNOWN || date == -1 || year == -1 || year > PR_INT16_MAX)
1093 if (year != -1)
1094 result->tm_year = year;
1159 * But if the year is 3001, mktime also invokes the invalid
1165 * zone, we consider year 3000 out of range as well. (See
