linux 内核中 mktime函数

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最新推荐文章于 2024-07-22 23:24:25 发布

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本文链接：https://blog.youkuaiyun.com/lgflgf/article/details/34849243

本文深入解析了Linux内核中mktime()函数的原理，详细阐述了计算当前时间距离1970-01-01的秒数的数学表达式，包括对月份天数的计算方法，以及不同月份情况下天数的特殊调整。通过公式推导，清晰展示了函数内部逻辑与计算过程。

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Linux内核中的mktime()函数.

计算当前时间(>=1970-01-01 00:00:00) 距离1970-01-01 00:00:00的秒数

具体定义如下：

unsigned long
mktime(const unsigned int year0, const unsigned int mon0,
const unsigned int day, const unsigned int hour,
const unsigned int min, const unsigned int sec)
{
unsigned int mon = mon0, year = year0;

/* 1..12 -> 11,12,1..10 */
if (0 >= (int) (mon -= 2)) {
mon += 12; /* Puts Feb last since it has leap day */
year -= 1;
}

return ((((unsigned long)
(year/4 - year/100 + year/400 + 367*mon/12 + day) +
year*365 - 719499
)*24 + hour /* now have hours */
)*60 + min /* now have minutes */
)*60 + sec; /* finally seconds */
}

函数实现的表达式看出，实现mktime 最关键的是找出计算每月天数的方法。

当前时间距离0001-01-01 的天数:

1. 当前时间的月份mon>2, D(mon > 2) = year/4-year/100+year/400 + (year-1) *365 + month_to_days(mon) + day-1

2. 当前时间的月份mon <=2, D(mon <= 2) = (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365+ month_to_days(mon) + day-1

现在问题在于如何求出month_to_days(mon)表达式.

假定2月份是30天, mon 与 month_to_days(mon) 关系如下:

mon days month_to_days(mon) 数列标记{Dm}

1 31 0 D1 = 0

2 30 31 D2 = 31

3 31 61 D3 = 61

4 30 92 D4 = 92

5 31 122 D5 = 122

6 30 153 D6 = 153

7 31 183 D7 = 183

8 31 214 D8 = 214

9 30 245 D9 = 245

10 31 275 D10 = 275

11 30 306 D11 = 306

12 31 336 D12 = 336

讨论: m <= 7 情况

D2 - D1 = 31

D3 - D2 = 30

D4 - D3 = 31

D5 - D4 = 30

D6 - D5 = 31

D7 - D6 = 30

D8 - D7 = 31

上式相加，可以得出: D2m+1 - D1 = 31*m+30*m = 61*m; D2m - D1 = 31*m+30*(m-1) = 61*m-30

由于D1 = 0. D2m+1 = 61*m ==> Dm = 61*(m-1)/2= (61*m-1)/2 - 30, m为<=7的奇数

D2m = 61*m-30 ==> Dm = 61*m/2 - 30 , m为<=7的偶数

讨论: m >7 情况

D9 - D8 = 31

D10 - D9 = 30

D11 - D10 = 31

D12 - D11 = 30

上式相加，可以得出: D2m+1 - D8 = 31*(m-3)+30*(m-4) = 61*m-213; D2m - D8 = 31*(m-4)+30*(m-4) = 61*m-244

由于D8 = 214， D2m+1 = 61*m + 1 ==>Dm = 61*(m-1)/2 + 1 = (61*m+1)/2 - 30, m为>7奇数

D2m = 61*m - 30 ==> Dm = 61*m/2 - 30 , m 为>7偶数

综上所述: Dm = 61*m/2 - 30, m为偶数

Dm = (61*m-1)/2 - 30, m为<=7 奇数

Dm = (61*m+1)/2 - 30, m为>7 奇数

上述是在假定2月份为30天得出的结论, 如果考虑m>2月份的情况, 此时月份与天的数关系为:

Dm = 61*m/2 - 30 - 61 + 31 + 28 = 61*(m-2)/2 + 29， m为3 <= m <= 12的偶数 (1)

Dm = (61*m-1)/2 - 30 - 61 +31 + 28 = (61*(m-2)-1)/2 + 29, m为3 <= m <= 7 奇数 (2)

Dm = (61*m+1)/2 - 30 - 61 + 31 + 28 = (61*(m-2)+1)/2 + 29, m为8 <= m <= 12 奇数 (3)

(3)式：8 <= m <= 12 ==> 6 <= m-2 <= 10

==> 1 <= (m-2)/6 <= 10/6

==> 61*(m-2)+1 <= 61*(m-2)+ (m-2)/6 <= 61*(m-2) + 10/6

==> (61*(m-2)+1)/2 <= (61*(m-2)+ (m-2)/6)/2 <= (61*(m-2) + 10/6)/2 = (61*(m-2) + 1)/2 + 1/3

由于 m为奇数, 所以 (61*(m-2) + 1)/2 为整数, 按计算机取整 (61*(m-2) + 1)/2 = (61*(m-2) + 1)/2 + 1/3

即 (61*(m-2)+1)/2 = (61*(m-2)+ (m-2)/6)/2 = 367*(m-2)/12

Dm = 367*(m-2)/12 + 29

(2)式: 3<= m <= 7 ==> 1 <= m-2 <= 5

==> -1 < 1/6 <= (m-2)/6 <= 5/6 < 1

==> 61*(m-2)-1 < 61*(m-2) + (m-2)/6 < 61*(m-2) +1

==> (61*(m-2)-1)/2 < (61*(m-2) + (m-2)/6)/2 < (61*(m-2) +1)/2

由于 m为奇数，所以 (61*(m-2)-1)/2 和 (61*(m-2) +1)/2 都为整数

而 (61*(m-2) +1)/2 -(61*(m-2)-1)/2 = 1

所以按计算机取整 (61*(m-2)-1)/2 = (61*(m-2) + (m-2)/6)/2 = 367*(m-2)/12

Dm = 367*(m-2)/12 + 29

(1)式: 3 <= m <= 12 ==> 1 <= m-2 <= 10

==> 0 < 1/12 <= (m-2)/12 <= 10/12 < 1

==> 61*(m-2)/2 < 61*(m-2)/2 + (m-2)/12 < 61*(m-2)/2 + 1

由m 为偶数, 所以 61*(m-2)/2 和 61*(m-2)/2 + 1 都为整数

而 61*(m-2)/2+1 - 61*(m-2)/2 = 1

所以按计算机取整 61*(m-2)/2 = 61*(m-2)/2 + (m-2)/12 = 367*(m-2)/12

Dm = 367*(m-2)/12 + 29

综上可以得出结论: m > 2 时, Dm = 367*(m-2)/12 + 29, 即 month_to_days(m) = 367*(m-2)/12 + 29

m = 1 时, Dm = 0, 即 month_to_days(1) = 0 = 367*(13-2)/12 + 29 - 365 = month_to_days(13) -365

m = 2时， Dm = 31, 即 month_to_days(2) = 31 = 367*(14-2)/12 + 29 - 365 = month_to_days(14) -365

即 m <= 2 时 month_to_days(m) = month_to_days(m+12) - 365

回到主问题:

当前时间距离0001-01-01 的天数:

1. 当前时间的月份mon>2, D(mon > 2) = year/4-year/100+year/400 + (year-1) *365 + month_to_days(mon) + day-1

2. 当前时间的月份mon <=2, D(mon <= 2) = (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365+ month_to_days(mon) + day-1

D(mon > 2) = year/4-year/100+year/400 + (year-1) *365 + month_to_days(mon) + day-1

= year/4-year/100+year/400 + (year-1) *365 + 367*(mon-2)/12 + 29+ day-1

= year/4-year/100+year/400 + year*365 + 367*(mon-2)/12 + day - 337

D(mon <= 2) = (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365+ month_to_days(mon) + day-1

= (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365 + month_to_days(mon+12) - 365 + day-1

= (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365 + 367*(mon+12-2)/12 + 29 - 365 + day-1

= (year-1)/4- (year-1)/100+ (year-1)/400 + (year-1)*365 + 367*(mon-2+12)/12 + day-337

D(1970-01-01 00:00:00) = 1969/4 - 1969/100 + 1969/400 + 1969*365 + 367*11/12 + 1 - 337 = 719162

D(mon > 2) - D(1970-01-01 00:00:00) = (year/4-year/100+year/400 + 367*(mon-2)/12 + day) + year*365- 719499

D(mon <= 2) - D(1970-01-01 00:00:00) = ((year-1)/4- (year-1)/100+ (year-1)/400 + 367*(mon-2+12)/12 + day) + (year-1)*365 -719499

即为linux mktime 函数实现计算距离1970-01-01 00:00:00 天数的表达式