Hello
On 20.01.14 12:31, sa9k063 wrote:
can someone please explain:
one of my boxes gets portscanned often by some likely infected laptops.
While having set
net.inet.tcp.blackhole=1
there are still messages like
+Limiting closed port RST response from 348 to 200 packets/sec
According to the blackhole(4) manpage (from a FreeBSD 9.1 system):
---8<------------------------------------------------------------
SYNOPSIS
sysctl net.inet.tcp.blackhole[=[0 | 1 | 2]]
sysctl net.inet.udp.blackhole[=[0 | 1]]
Part of DESCRIPTION:
Normal behaviour, when a TCP SYN segment is received on a port
where there is no socket accepting connections, is for the system
to return a RST segment, and drop the connection. The connecting
system will see this as a “Connection refused”. By setting the
TCP blackhole MIB to a numeric value of one, the incoming SYN
segment is merely dropped, and no RST is sent, making the system
appear as a blackhole. By setting the MIB value to two, any
segment arriving on a closed port is dropped without returning a
RST. This provides some degree of protection against stealth
port scans.
---8<------------------------------------------------------------
So it is possible, that you are hit with something else then SYN
packets and should probably set net.inet.tcp.blackhole=2, or even
with UDP packets, then also set net.inet.udp.blackhole=1.
What output does 'sysctl -a | grep blackhole' show?
bye
Fabian
_______________________________________________
[email protected] mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-security
To unsubscribe, send any mail to "[email protected]"
