linux内核计算从1970年1月1日0时起到开机当日经过的秒数，采用的方法不是调用开发环境库中的函数，
而是linus专门实现的一个。个人觉得在处理闰年问题时很经典，特此贴出来为日后留个纪念。
long kernel_mktime(struct tm * tm) |
{ |
long res; |
int year; |
|
year = tm->tm_year - 70; |
/* magic offsets (y+1) needed to get leapyears right.*/ |
res = YEAR*year + DAY*((year+1)/4); |
res += month[tm->tm_mon]; |
/* and (y+2) here. If it wasn't a leap-year, we have to adjust */ |
if (tm->tm_mon>1 && ((year+2)%4)) |
res -= DAY; |
res += DAY*(tm->tm_mday-1); |
res += HOUR*tm->tm_hour; |
res += MINUTE*tm->tm_min; |
res += tm->tm_sec; |
return res; |
} |
注解：1.(year+1)/4被称作一个魔幻值，用来计算1970年以来的闰年数，而(year+2)%4则是用来判断是不是闰年. |
#define MINUTE 60 |
#define HOUR (60*MINUTE) |
#define DAY (24*HOUR) |
#define YEAR (365*DAY) |
|
/* interestingly, we assume leap-years */ |
static int month[12] = { |
0, |
DAY*(31), |
DAY*(31+29), |
DAY*(31+29+31), |
DAY*(31+29+31+30), |
DAY*(31+29+31+30+31), |
DAY*(31+29+31+30+31+30), |
DAY*(31+29+31+30+31+30+31), |
DAY*(31+29+31+30+31+30+31+31), |
DAY*(31+29+31+30+31+30+31+31+30), |
DAY*(31+29+31+30+31+30+31+31+30+31), |
DAY*(31+29+31+30+31+30+31+31+30+31+30) |
}; |