學(xué)會用Linux0.11-啟動時間的計算函數(shù)

main.c下時間初始化代碼，可以體會到位運算的精簡

#define CMOS_READ(addr) ({ /

outb_p(0x80|addr,0x70); /

inb_p(0x71); /

})

#define BCD_TO_BIN(val) ((val)=((val)&15) + ((val)>>4)*10)

static void time_init(void)

{

struct tm time;

do {

time.tm_sec = CMOS_READ(0);

time.tm_min = CMOS_READ(2);

time.tm_hour = CMOS_READ(4);

time.tm_mday = CMOS_READ(7);

time.tm_mon = CMOS_READ(8);

time.tm_year = CMOS_READ(9);

} while (time.tm_sec != CMOS_READ(0));

BCD_TO_BIN(time.tm_sec);

BCD_TO_BIN(time.tm_min);

BCD_TO_BIN(time.tm_hour);

BCD_TO_BIN(time.tm_mday);

BCD_TO_BIN(time.tm_mon);

BCD_TO_BIN(time.tm_year);

time.tm_mon--;

startup_time = kernel_mktime(&time);

}

#define CMOS_READ(addr) ({ /outb_p(0x80|addr,0x70); /inb_p(0x71); /})#define BCD_TO_BIN(val) ((val)=((val)&15) + ((val)>>4)*10)static void time_init(void){struct tm time;do {time.tm_sec = CMOS_READ(0);time.tm_min = CMOS_READ(2);time.tm_hour = CMOS_READ(4);time.tm_mday = CMOS_READ(7);time.tm_mon = CMOS_READ(8);time.tm_year = CMOS_READ(9);} while (time.tm_sec != CMOS_READ(0));BCD_TO_BIN(time.tm_sec);BCD_TO_BIN(time.tm_min);BCD_TO_BIN(time.tm_hour);BCD_TO_BIN(time.tm_mday);BCD_TO_BIN(time.tm_mon);BCD_TO_BIN(time.tm_year);time.tm_mon--;startup_time = kernel_mktime(&time);}

kernel/mktime.c代碼，其中包含了閏年的計算，但是代碼確實那么精簡，這就是算法和思想

閏年的基本計算方法是:

如果 y 能被 4 除盡且不能被 100 除盡,或者能被 400 除盡,則 y 是閏年。

/*

* linux/kernel/mktime.c

*

* (C) 1991 Linus Torvalds

*/

#include

/*

* This isn't the library routine, it is only used in the kernel.

* as such, we don't care about years<1970 etc, but assume everything

* is ok. Similarly, TZ etc is happily ignored. We just do everything

* as easily as possible. Let's find something public for the library

* routines (although I think minix times is public).

*/

/*

* PS. I hate whoever though up the year 1970 - couldn't they have gotten

* a leap-year instead? I also hate Gregorius, pope or no. I'm grumpy.

*/

#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)

};

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;

}

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