return (time_t) -1;
days = 365 * (t->tm_year - 70);
- /* Take into account the leap days between 1970 and YEAR-1; all
- years divisible by four between 1968 and 2100 should be leap. */
+ /* Take into account leap years between 1970 and YEAR-1; all years
+ divisible by four between 1968 and 2100 should be leap. */
days += (t->tm_year - 1 - 68) / 4;
if (t->tm_mon < 0 || t->tm_mon >= 12)
return (time_t) -1;