1
0
mirror of https://github.com/rkd77/elinks.git synced 2024-12-04 14:46:47 -05:00
elinks/src/protocol
Kalle Olavi Niemitalo b7d3b4f687 1041: Add ftp_add_unparsed_line: HTML entities and more error checks.
Separate the formatting of unparsed lines from ftp_process_dirlist()
to a new function ftp_add_unparsed_line().  Check for all possible
out-of-memory errors.  Encode HTML metacharacters as entity references
and document how charsets are handled FTP directory listings.
Add a NEWS entry.
2008-09-04 11:21:06 +03:00
..
auth Strings corrections from Malcolm Parsons 2008-01-27 04:19:23 +00:00
bittorrent Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
file Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
finger Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
fsp Work around fsp_open_session() not setting errno. 2008-08-03 20:26:50 +03:00
ftp 1041: Add ftp_add_unparsed_line: HTML entities and more error checks. 2008-09-04 11:21:06 +03:00
gopher Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
http Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
nntp Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
rewrite Minor fixes and improvements to option strings 2008-03-06 10:31:20 +02:00
smb Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
test Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
about.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
about.h
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
date.c parse_time: set tm_sec to zero before conditional second parsing 2008-02-28 23:38:42 +02:00
date.h
header.c
header.h
Makefile Bug 744: Add tests. There are four failures. 2007-07-19 13:46:47 +03:00
protocol.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
protocol.h Bug 744: Make removal of double slashes more protocol specific 2007-09-11 14:14:17 +02:00
proxy.c Bug 1013: Don't assume errno is between 0 and 100000 2008-08-03 17:56:41 +03:00
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 bug 1000: Do not discard the query part of URI. 2008-03-02 17:45:29 +02:00
uri.h Move is_in_domain from cookies/cookies.c to protocol/uri.c and export 2007-09-14 16:51:04 +02:00
user.c Trim trailing whitespaces. 2007-09-14 15:12:32 +02:00
user.h

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