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* given a struct sock, how to find pid of process that owns it?
@ 2003-09-03 17:17 Chris Friesen
  2003-09-03 19:00 ` Stephen Hemminger
  0 siblings, 1 reply; 2+ messages in thread
From: Chris Friesen @ 2003-09-03 17:17 UTC (permalink / raw)
  To: Linux Kernel Mailing List, netdev


I'm working on a small app similar to netstat that only cares about unix 
sockets.

I can easily walk /proc/net/unix, but to find the owner of the socket I 
need to scan /proc, which gets expensive.

Accordingly, I'd like to extend /proc/net/unix to also dump out the pid 
of the process that owns the socket.  The only thing is, I can't seem to 
figure out how to find the pid of the socket owner given a pointer to 
the socket struct.

Any tips?  Is it even there?

Thanks,

Chris


-- 
Chris Friesen                    | MailStop: 043/33/F10
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^ permalink raw reply	[flat|nested] 2+ messages in thread
* Re: given a struct sock, how to find pid of process that owns it?
  2003-09-03 17:17 given a struct sock, how to find pid of process that owns it? Chris Friesen
@ 2003-09-03 19:00 ` Stephen Hemminger
  0 siblings, 0 replies; 2+ messages in thread
From: Stephen Hemminger @ 2003-09-03 19:00 UTC (permalink / raw)
  To: linux-kernel

On Wed, 03 Sep 2003 13:17:25 -0400
Chris Friesen <cfriesen@nortelnetworks.com> wrote:

> 
> I'm working on a small app similar to netstat that only cares about unix 
> sockets.
> 
> I can easily walk /proc/net/unix, but to find the owner of the socket I 
> need to scan /proc, which gets expensive.
> 
> Accordingly, I'd like to extend /proc/net/unix to also dump out the pid 
> of the process that owns the socket.  The only thing is, I can't seem to 
> figure out how to find the pid of the socket owner given a pointer to 
> the socket struct.
> 

There is a N to 1 relationship, you will end up needing the scan.

^ permalink raw reply	[flat|nested] 2+ messages in thread
end of thread, other threads:[~2003-09-03 19:04 UTC | newest]

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2003-09-03 17:17 given a struct sock, how to find pid of process that owns it? Chris Friesen
2003-09-03 19:00 ` Stephen Hemminger

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