题
这似乎喜欢的东西没有人应该要做,但是我工作的一个核心模块,用于嵌入式系统(本项目)在这似乎 time.h
做 包括 timespec
和 time_t
类型,以及 clock_gettime
和 gmtime
功能,但并 不 包括 localtime
, ctime
, time
, 或者,极度, tm
类型。
当我试图投返回的指针,从gmtime到我自己的结构,我获得一个出现段错误.
所以我想我能内容,以解决该问题的两个方法—这会是伟大的图找出如何获得失的类型,或者,如何推我自己的方法解unix时间戳。
解决方案
此应该是准确的(填写了struct tm
的切口向下仿,我的year
使用共同的时代,而不是1900 CE历元):
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。
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