Marvin George wrote:
> At 01:14 PM 11/6/98 -0500, Edward Doolittle wrote:
>> On Fri, 6 Nov 1998, Marvin George wrote:
>>
>>> For over 9 months, we have successfully used Linux (RH 5.0) with diald
>>> and IP masquerading to put our LAN on the Internet. Our former ISP used
>>> a version of Unix (unknown which one). We recently switched to a
>>> different ISP who uses (UGH!!!) NT. The dynamically assigned local IP
>>> is not present in the output of ifconfig -- it is using the same IP as
>>> the ethernet card, which is from the private set of network numbers
>>> (192.168.1.x). When I log on using dip, it works correctly, assigning
>>> the dynamic IP to the local (ppp0) interface. With both dip and diald,
>>> the remote IP is correctly assigned.
>>> Also, when logging on with dip, IP masquerading does not work.
>>
>> Odd. Where is your IP masquerading set up? Please post your
>> configuration files, output from ifconfig, ipfwadm, etc. so we can see the
>> details.
>
> See the end of this message. (Sorry about the length....)
We wanted it to be long. Longer than you had, even. /etc/diald.conf
seems to be missing. Course, the connect script was a little
gratuitously long...
> Internet. Why does dip pick up the new ISP's dynamically-assigned inet
> address, but diald doesn't? And why does diald pick it up from the old ISP
> but not the new one? Both dip and diald pick up the P-t-P address OK,
> regardless of which ISP is being connected to. I think that, when the inet
> address is wrong but the P-t-P address is OK, our Linux box sends out
> packets OK but does not respond to received packets at all.
No, it doesn't receive packets. Your network you claim to send from is
a private network address (as it should be, for strictly internal
networks), and so the ISP sees it, and says, "hey, this shouldn't be
here.", and chucks it. As it should. Yeah, it's inconvinient when you
have problems like yours, but that just makes you fix it. If it didn't
work that way, you'd have interrupted internet accessibility to probably
dozens of people, and most likely through little fault of your own.
> Here's the output of ifconfig (edited to remove some eth0:? aliases) with
> diald running, not connected:
>
> eth0 Link encap:Ethernet HWaddr 00:20:18:25:18:C9
> inet addr:192.168.1.57 Bcast:192.168.1.255 Mask:255.255.255.0
>
> sl0 Link encap:Serial Line IP
> inet addr:192.168.1.57 P-t-P:192.168.0.2 Mask:255.255.255.0
Don't do this. No. No. No.
Every interface on your machine should have a unique IP address. Why?
It works more reliably that way. Many people use a single address on
multiple interfaces, and have no problems. Many other people try it and
it breaks bad. Do not do this if you do not like taking risks. Since I
have yet to find something which works better when you have multiple
interfaces with the same IP, I'll tell you again, NO. (I do know of
some 'neat hacks' which can be done. However, they generally work
better if you do it right...)
> # and set the rule for masquerading from local net
> ipfwadm -F -a masquerade -P all -S 192.168.0.0/16 -D 0.0.0.0/0
> ipfwadm -F -a masquerade -P all -S 209.181.112.0/24 -D 0.0.0.0/0
> =============================
> NOTE: The '-S 209.---' gets changed as appropriate for the ISP being
> connected to; for some reason, the above rules don't seem to work with the
> NT-based ISP, even with the correct net address in place. Nor does it work
> with the Unix-based ISP when I connect via dip.
um, the -S 209... is, um, wrong. Why are you trying to masquerade your
ISP and its perfectly valid IPs? The first masquerade rule only,
please. (Though, that shouldn't cause any major problems, since your
ISP won't route through you. Err, maybe it will. *ponder*)
Other than this, I can't see anything definitely wrong. Not sure how
well diald-top runs from the connect script, though. When you think
about how it can... Oh, it was just a netscape or pine line wrap? Ok.
I've ranted once this message, so I'll not say anything abou. No, I
won't say anything. Sorry.
Hopefully, I will have a better answer than I did, or maybe someone else
will answer.
Ed
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