message: Simplify day-of-year calculation, correct comment
diff: @@ -250,15 +250,13 @@ pub unsafe extern "C" fn gmtime_r(clock: *const time_t, result: *mut tm) -> *mut
  * 1900) */
  let is_leap_year: bool = year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
  - /* For dates that are March 1 or later, we can use day-of-
  - * year in the transformed calendar. For January and
  - * February, that value is sensitive to whether the previous
  - * year is a leap year. Therefore, we use the already
  - * computed date for those two months. */
  - let yday: c_int = match month {
  - 0 => mday - 1, // January
  - 1 => 31 + mday - 1, // February
  - _ => day_of_year_transformed + if is_leap_year { 60 } else { 59 },
  + /* For dates in January or February, we use the fact that
  + * January 1 is always 306 days after March 1 in the
  + * previous year. */
  + let yday: c_int = if month < 2 {
  + day_of_year_transformed - 306
  + } else {
  + day_of_year_transformed + if is_leap_year { 60 } else { 59 }
  };
  let hour: c_int = (secs_of_day / (60 * 60)).try_into().unwrap();
  