1
0
mirror of https://github.com/rkd77/elinks.git synced 2024-09-20 01:46:15 -04:00
elinks/src/protocol
Witold Filipczyk 7453b03279 deflate switched on again.
askubuntu.com uses deflate compression.
2014-09-01 12:50:23 +02:00
..
auth Memorize the auth form only when the "document.browse.forms.show_formhist" is on. 2011-10-06 10:54:11 +02:00
bittorrent doxygen: Escape <uri> in doc of bittorrent_download 2012-11-18 20:16:59 +02:00
file bug 764: Initialize the right member of union option_value 2012-11-03 22:16:32 +02:00
finger Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
fsp Merge branch 'elinks-0.12' into elinks-0.13 2008-08-03 22:18:53 +03:00
ftp bug 764: Initialize the right member of union option_value 2012-11-03 22:16:32 +02:00
gopher Merge branch 'elinks-0.12' into elinks-0.13 2008-11-01 22:39:17 +02:00
http deflate switched on again. 2014-09-01 12:50:23 +02:00
nntp bug 764: Initialize the right member of union option_value 2012-11-03 22:16:32 +02:00
rewrite rewrite: update default dumb and smart prefixes 2012-11-30 09:24:53 +02:00
smb SMB directory listing looks like normal directory listing. 2010-09-06 18:38:36 +02:00
test Compilation fix. make test failed. 2010-08-08 09:38:39 +02:00
about.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
about.h Remove empty lines in start of header files 2005-11-15 11:33:27 +01:00
common.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
common.h Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
data.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
data.h Remove empty lines in start of header files 2005-11-15 11:33:27 +01:00
date.c parse_time: set tm_sec to zero before conditional second parsing 2008-02-28 23:38:42 +02:00
date.h Remove empty lines in start of header files 2005-11-15 11:33:27 +01:00
header.c parse_header: document parameters and return value 2012-11-18 17:36:36 +02:00
header.h parse_header: make the 2nd parameter point to const 2012-11-18 16:39:04 +02:00
Makefile Bug 744: Add tests. There are four failures. 2007-07-19 13:46:47 +03:00
protocol.c bug 764: Initialize the right member of union option_value 2012-11-03 22:16:32 +02:00
protocol.h Handle mailcap's copiousoutput without an external pager. 2010-07-24 17:07:18 +02:00
proxy.c Merge branch 'elinks-0.12' into elinks-0.13 2008-11-01 22:39:17 +02:00
proxy.h Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
README.timegm Move README.timegm to the same directory as date.c. 2007-01-13 10:01:51 +02:00
uri.c Revert "strcpy -> strlcpy." 2010-09-19 15:26:05 +02:00
uri.h 1008: percent-encode file names in uri.post 2008-07-11 14:44:35 +03:00
user.c bug 764: Initialize the right member of union option_value 2012-11-03 22:16:32 +02:00
user.h Remove empty lines in start of header files 2005-11-15 11:33:27 +01:00

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<53>phane