Messages in this thread | | | Date | Tue, 16 Sep 2003 13:21:34 -0700 | From | Vishwas Raman <> | Subject | Re: Incremental update of TCP Checksum |
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Richard B. Johnson wrote: > On Tue, 16 Sep 2003, Vishwas Raman wrote: > > >>Hi all, >> >>I have a very simple question, which a lot of you would have solved. I >>am intercepting a TCP packet, which I would like to change slightly. >> >>Let's say, I change the doff field of the tcp-header (for eg: increase >>it by 1). I know it is wrong just to change the doff field without >>increasing the packet length, but lets say I do it just as a test. Since >>I changed a portion of the tcp header, I have to update the tcp checksum >>too right!!! If so, what is the best way to do so, without having to >>recalculate the entire tcp checksum (I know how to recalculate the >>checksum from scratch). >> >>Can anyone out there tell me the algorithm to update the checksum >>without having to recalculate it. >> >>I tried the following algorithm but it didnt work. The packet got >>rejected as a packet with bad cksum. >> >>void changePacket(struct sk_buff* skb) >>{ >> struct tcphdr *tcpHdr = skb->h.th; >> // Verifying the tcp checksum works here... >> tcpHeader->doff += 1; >> long cksum = (~(tcpHdr->check))&0xffff; >> cksum += 1; >> while (cksum >> 16) >> { >> cksum = (cksum & 0xffff) + (cksum >> 16); >> } >> tcpHeader->check = ~cksum; >> // Verifying tcp checksum here fails with bad cksum >>} >> >>Any pointers/help in this regard will be highly appreciated... > > > The TCP/IP checksum is a WORD sum (unsigned short) in which > any overflow out of the word causes the word to be incremented. > The final sum is then inverted to become the checksum. Note that > many algorithms sum into a long then fold-back the bits. It's > the same thing, different method. > > Therefore: > Given an existing checksum of 0xffff, if the > next word to be summed is 0x0001, the result > will be 0x0001 because adding 1 to 0xffff makes > it 0, causing an overflow which propagates to > become 0x0001. > So: > Clearly, information is lost because one doesn't > know how the 0x0001 was obtained. > > If I were to modify a low byte somewhere by subtracting 1, > would I know that the new checksum, excluding the inversion, > was 0x0000? No. It could be 0xffff. > > This presents a problem when trying to modify existing checksums. > It's certainly easier to set the existing checksum to 0, then > re-checksum the whole packet. It's probably faster than some > looping algorithm that attempts to unwind a previous checksum.
Are you then suggesting that instead of trying to do an incremental update of the tcp checksum, I set it to 0 and recalculate it from scratch? But I thought that doing that was a big performance hit. Isn't it?
> > > Cheers, > Dick Johnson > Penguin : Linux version 2.4.22 on an i686 machine (794.73 BogoMips). > Note 96.31% of all statistics are fiction. > > >
-- -- Vishwas Raman Software Engineer, Eternal Systems, Inc, 5290 Overpass Rd, Bldg D, Santa Barbara. CA 93111 Email: vishwas@eternal-systems.com Tel: (805) 696-9051 x246 Fax: (805) 696-9083 URL: http://www.eternal-systems.com/
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