这应该是准确的(填写 a 的缩减模仿struct tm
,我year
使用 Common Era 而不是 1900 CE epoch):
struct xtm
{
unsigned int year, mon, day, hour, min, sec;
};
#define YEAR_TO_DAYS(y) ((y)*365 + (y)/4 - (y)/100 + (y)/400)
void untime(unsigned long unixtime, struct xtm *tm)
{
/* First take out the hour/minutes/seconds - this part is easy. */
tm->sec = unixtime % 60;
unixtime /= 60;
tm->min = unixtime % 60;
unixtime /= 60;
tm->hour = unixtime % 24;
unixtime /= 24;
/* unixtime is now days since 01/01/1970 UTC
* Rebaseline to the Common Era */
unixtime += 719499;
/* Roll forward looking for the year. This could be done more efficiently
* but this will do. We have to start at 1969 because the year we calculate here
* runs from March - so January and February 1970 will come out as 1969 here.
*/
for (tm->year = 1969; unixtime > YEAR_TO_DAYS(tm->year + 1) + 30; tm->year++)
;
/* OK we have our "year", so subtract off the days accounted for by full years. */
unixtime -= YEAR_TO_DAYS(tm->year);
/* unixtime is now number of days we are into the year (remembering that March 1
* is the first day of the "year" still). */
/* Roll forward looking for the month. 1 = March through to 12 = February. */
for (tm->mon = 1; tm->mon < 12 && unixtime > 367*(tm->mon+1)/12; tm->mon++)
;
/* Subtract off the days accounted for by full months */
unixtime -= 367*tm->mon/12;
/* unixtime is now number of days we are into the month */
/* Adjust the month/year so that 1 = January, and years start where we
* usually expect them to. */
tm->mon += 2;
if (tm->mon > 12)
{
tm->mon -= 12;
tm->year++;
}
tm->day = unixtime;
}
我为所有神奇的数字道歉。367*month/12 是生成日历的 30/31 天序列的巧妙技巧。该计算适用于从 3 月开始直到最后修正的年份,这使事情变得容易,因为闰日落在“年”的末尾。