mktime.c中好像存在问题

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这是0.11内核中的一个文件
/*
* 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;
}

看第一遍时觉得kernel_mktime()

函数写的很是巧妙，再仔细琢磨时，个人觉得这个函数好像有点问题(也许是个人能力有限，不是函数有问题，而是自己没有领悟到更深的一层)。
在kernel/mktime.c 中的kernel_mktime()函数原型如下：
    41    long kernel_mktime(struct tm * tm)
    42    {
    43        long res;
    44        int year;
    45
    46        year = tm->tm_year - 70;
    47    /* magic offsets (y+1) needed to get leapyears right.*/
    48        res = YEAR*year + DAY*((year+1)/4);
    49        res += month[tm->tm_mon];
    50    /* and (y+2) here. If it wasn't a leap-year, we have to adjust */
    51        if (tm->tm_mon>1 && ((year+2)%4))
    52            res -= DAY;
    53        res += DAY*(tm->tm_mday-1);
    54        res += HOUR*tm->tm_hour;
    55        res += MINUTE*tm->tm_min;
    56        res += tm->tm_sec;
    57        return res;
    58    }
我的问题处在第48行的“ ((year+1)/4)“上，这行代码功能：计算从1970年到现在经历的年数，再将计算出来的年数乘以一年的秒数，转化为以秒为单位的时间。
((year+1)/4)是计算从1970年到现在的闰年数，他是按照闰年没4年出现一次的规律计算的。
而闰年是每4年出现一次吗？不是的。400年中闰年只有97个，而不是100个，1900不是闰年，1800不是，2000是。
所以他这样计算出来的闰年数对吗？

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