Linux 0.11启动时间的计算函数

最新推荐文章于 2023-03-28 10:36:30 发布

最新推荐文章于 2023-03-28 10:36:30 发布 · 68 阅读

本文介绍了一个在main.c文件中实现的时间初始化函数，该函数通过位运算读取CMOS时钟并将其从BCD码转换为十进制数。此外，还详细解释了kernel/mktime.c中的时间转换函数，它将结构化时间转换为自1970年以来的秒数，并考虑了闰年的计算。

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main.c下时间初始化代码，可以体会到位运算的精简

#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代码，其中包含了闰年的计算，但是代码确实那么精简，这就是算法和思想

闰年的基本计算方法是:

如果 y 能被 4 除尽且不能被 100 除尽,或者能被 400 除尽,则 y 是闰年。

/* * linux/kernel/mktime.c * * (C) 1991 Linus Torvalds */ #include <time.h> /* * 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; }