Re: Why are there two sshd processes when I log into my computer?

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Knute Johnson wrote:
>> When I log in remotely to my FC5 box two sshd processes with usually 
>> consecutive pids are created?  The original sshd process is still 
>> there too.
> 
> Any ideas then why one of them logs its time in UTC and the other 
> local?
> 
> May 24 22:43:13 rabbitbrush sshd[4258]: Accepted publickey for knute 
> from 208.1.40.46 port 1614 ssh2
> May 25 05:43:13 rabbitbrush sshd[4259]: Accepted publickey for knute 
> from 208.1.40.46 port 1614 ssh2

I agree that it is annoying to have incorrect logs.

This problem has been raised before:

  http://www.redhat.com/archives/rhl-list/2006-March/msg03980.html
  http://www.redhat.com/archives/rhl-list/2006-April/msg00778.html

but I don't see any reply about a solution and I didn't find an
open bug for it.

It looks like environmental variables are discarded for security
reasons (try running "cat /proc/4258/environ", with the correct pid;
there is no environment). I don't know if the two facts can be related.

I just tried "strace -f" on the sshd daemon process: the following
lines appear interesting

[pid  9996] time(NULL)                  = 1148562034
[pid  9996] open("/etc/localtime", O_RDONLY) = -1 ENOENT (No such file or directory)
[pid  9996] open("/etc/localtime", O_RDONLY) = -1 ENOENT (No such file or directory)
[pid  9996] open("/etc/localtime", O_RDONLY) = -1 ENOENT (No such file or directory)
[pid  9996] open("/etc/localtime", O_RDONLY) = -1 ENOENT (No such file or directory)
[pid  9996] send(6, "<86>May 25 13:00:34 sshd[9996]: "..., 94, MSG_NOSIGNAL) = 94

The process (which is running as user "sshd") does not see the
"/etc/localtime" file and logs a wrong time (13:00 instead of 15:00,
in this case).

I don't know why the file appears totally invisible to the process
(chroot tricks?).

Best regards.
-- 
   Roberto Ragusa    mail at robertoragusa.it

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