On Sun, Oct 21, 2007 at 10:53:29PM +0200, Tal Kelrich wrote:
> On Sun, 21 Oct 2007 19:29:02 +0200
> Willy Tarreau <[email protected]> wrote:
>
> > Hi,
> >
> > On Sun, Oct 21, 2007 at 05:53:37PM +0200, Tal Kelrich wrote:
> > > Hi,
> > >
> > > I've run into a problem where a process trying to connect to a local
> > > port within the local port range eventually ends up connected to
> > > itself, with source port = dest port.
> > >
> > > similar behavior can be gotten by running netcat as follows:
> > > nc -p 1025 localhost 1025
> > >
> > > I'm not really sure if that's a bug, but the original case was at
> > > least unexpected.
> >
> > It is not a bug, it is caused by the "simultaneous connect" feature of
> > TCP. Although rarely used, in TCP you can connect two clients
> > together. They just have to exchange their SYN, SYN/ACK then ACK and
> > bingo, they're connected. In fact, you found the easiest way to
> > achieve it, by using the same port. To demonstrate the feature, I'm
> > used to either temporarily block SYNs with iptables, or by unplugging
> > the cable between two machines.
> >
>
> Hi,
>
> It still seems confusing that a connect against localhost may
> randomly succeed.
>
> Here's a better example, if somewhat violent. eventually succeeds.
> (p=$((`cat /proc/sys/net/ipv4/ip_local_port_range|cut -f1`+1)); while
> true ; do nc -v -v 127.0.0.1 $p ; done)
Of course, for the same reason. If you reduce the ip_local_port_range, it
will even succeed more often. This is because the source port is choosen
before the first packet is sent, so when it is sent, it reaches a pending
connection (itself).
I can understand that it is confusing when you see it as a single
connection, but try to imagine (or reproduce) with 2 machines, then
translate that to the localhost with a single and same connection.
You may even draw the exchanges on paper, an you will notice that
"each end" of the connection gets its SYN-SYN/ACK-ACK sequence.
You may also tcpdump on loopback if that helps.
Regards,
Willy
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