如何分解UNIX时间用C(How to decompose unix time in C)

这看起来似乎任何人都不应该要做的，但我工作在它看来，嵌入式系统（的OpenWRT）内核模块time.h 不包括timespec和time_t类型，和clock_gettime和gmtime功能，但不包括localtime ， ctime ， time ，或，关键的是， tm型。

当我试图从gmtime的返回指针转换为我自己的结构，我得到一个段错误。

所以我想我会满足于解决问题的两种方法之一，it'd是巨大的，弄清楚如何获得访问该缺少类型，或者，如何推出自己的方法分解unix时间戳。

Answer 1:

这应该是准确的（填写一个的删节仿struct tm ，我的year使用共同的时代，而不是1900 CE纪元）：

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;
}

我的道歉对于所有的幻数。 367 *月/ 12是一个巧妙的方法来产生日历的30/31天序列。计算可与年中开始在三月直到修正到了最后，这让事情变得容易，因为那时的闰日落在一个“年”的结束。