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. -- Stéphane