readme.timegm

来自「elinks下lynx是最重要的二个文本浏览器, 在linux下非常实用, el」· TIMEGM 代码 · 共 65 行

This file contains description of our my_timegm() function in date.c. It was
posted as a mail to links-list by Stephane Chazelas, and I thought it may be
interesting for someone, so I decided to include it in the ELinks distribution.




Un explanation for it as one (me to start with) may wonder why
this works.

We first change of calendar. To make things easy, let's say
that 0/0/0 0:0:0 in our new calendar is the Marsh 1st 1968, so
just after a february 29th as 1968 was a leap year.

if y/m/d h:min:s is time in our calendar
and
   Y/M/D h:min:s in the new calendar

M = m - 1 - 2 (+ 12 if m < 3)
Y = y - 68 (-1 if m < 3)
D = d - 1

at Y/0/0 0:0:0, there has been Y / 4 leap years in the past, so
(int) 365 * Y + Y / 4 days have past.

at Y/M/0 0:0:0, lets find how many days have past since Y/0/0:

                   |Mar                                        Feb
                  M| 0   1   2   3   4   5   6   7   8   9  10  11
-------------------+-----------------------------------------------
 days in that month|31  30  31  30  31  31  30  31  30  31  31  28 or 29
-------------------+-----------------------------------------------
  x = days at Y/M/0| 0  31  61  92 122 153 184 214 245 275 306 337
-------------------+-----------------------------------------------
first approximation|
         y = 30 * M| 0  30  60  90 120 150 180 210 240 270 300 330
-------------------+-----------------------------------------------
              x - y| 0   1   1   2   2   3   4   4   5   5   6   7
-------------------+-----------------------------------------------
(M + 4) * 3 / 5 - 2| 0   1   1   2   2   3   4   4   5   5   6   7
-------------------+-----------------------------------------------

x - y = (M + 4) * 3 / 5 - 2

x = 30 * M + (M + 4) * 3 / 5 - 2

x = (153 * M + 2) / 5

So at Y/M/D 0:0:0,

Y * 1461 / 4 + (153 * M + 2) / 5 + D days have past since
the 1st of March of 1968

1st of March of 1968 was 671 days before 1970 Jan 1st (year 0
for unix time())

So
t = s + 60 * (min + 60 * (h + 24 * (Y * 1461 / 4 + (153 * M + 2) / 5 + D - 671)))
t = s + 60 * (min + 60 * (h + 24 * (Y * 1461 / 4 + (153 * M + 2) / 5 + d - 672)))

This shouldn't work past 2100/02/28 23:59:59 as 2100 is not a leap year.

--
St閜hane