--On 10 July 2004 15:31 +1000 Zeus Ng <zeus.ng(a)isquare.com.au> wrote:
if (method=="INVITE") {
if (isflagset(4) && isflagset(5)) { /* NAT source & NAT dest. */
if (!exec_msg('echo $SIP_RURI | cut -d: -f2 | cut -d@ -f2 | grep
"$SIP_SRCIP" > /dev/null')) {
xlog("L_NOTICE", "UAs behind different NAT devices, forcing
rtpproxy\n");
force_rtp_proxy();
t_on_reply("2");
} else {
xlog("L_NOTICE", "UAs behind same NAT devices\n");
t_on_reply("3");
}
... Other situation, NAT -> Public, Public -> NAT ...
}
For the sake of pedantry, I think that test will give a false positive
(i.e. wrongly treat caller and callee as being behind the same NAT)
in situations similar to:
10.0.0.1 A -> NAT1 -192.168.0.1 \
|--> NAT3 --> 195.1.1.1 Internet
10.0.0.2 B -> NAT2 -192.168.0.2 /
Alex