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C 编程 TCP 校验和

[英]C Programming TCP Checksum

I have been having trouble doing the checksum for TCP for several days now.几天来,我一直在为 TCP 进行校验和处理时遇到问题。 I have looked at many sources on the Internet but none of the examples that I have seen show you how to do the TCP checksum.我查看了 Internet 上的许多资源,但我所看到的示例都没有向您展示如何进行 TCP 校验和。 I have also looked at the RFC document and still I am having trouble:我还查看了RFC文档,但仍然遇到问题:

Below is the code I am using to generate the checksum:下面是我用来生成校验和的代码:

unsigned short checksum(unsigned short * buffer, int bytes)
{
    unsigned long sum = 0;
    unsigned short answer = 0;
    int i = bytes;
    while(i>0)
    {
            sum+=*buffer;
            buffer+=1;
            i-=2;
    }
    sum = (sum >> 16) + (sum & htonl(0x0000ffff));
    sum += (sum >> 16);
    return ~sum;
}

This function works for the IP checksum.此功能适用于 IP 校验和。

Below is the struct I have made for my TCP header:下面是我为我的 TCP 标头制作的结构:

struct tcp_header
{
    unsigned short tcp_sprt;
    unsigned short tcp_dprt;
    unsigned int tcp_seq;
    unsigned int tcp_ack;
    unsigned char tcp_res:4;
    unsigned char tcp_off:4;
    unsigned char tcp_flags;
    unsigned short tcp_win;
    unsigned short tcp_csum;
    unsigned short tcp_urp;
};

I have been using Wireshark to test these packets and the only thing wrong is the checksum.我一直在使用 Wireshark 来测试这些数据包,唯一错误的是校验和。

Finally here is the pseudo header struct that I load up with the TCP header and information from the IP header:最后,这是我用 TCP 标头和来自 IP 标头的信息加载的伪标头结构:

struct pseudoTcpHeader
{
    unsigned int ip_src;
    unsigned int ip_dst;
    unsigned char zero;//always zero
    unsigned char protocol;// = 6;//for tcp
    unsigned short tcp_len;
    struct tcp_header tcph;
};

Once I load up this struct with the correct information I then use the checksum function on the entire pseudo header struct and assign the TCP checksum to that value.一旦我用正确的信息加载了这个结构,我就会在整个伪头结构上使用校验和函数,并将 TCP 校验和分配给该值。 Do you see anything wrong with what I have provided?你看到我提供的东西有什么问题吗? If the problem isn't here it may be a careless error that I can't see.如果问题不在这里,那可能是我看不到的粗心错误。

I found a fairly good example on the winpcap-users mailing list which should address Greg's comment about odd length data and give you something to compare your code against.我在winpcap-users 邮件列表上找到了一个相当不错的例子,它应该解决 Greg 关于奇数长度数据的评论,并为您提供一些比较您的代码的内容。

USHORT CheckSum(USHORT *buffer, int size)
{
    unsigned long cksum=0;
    while(size >1)
    {
        cksum+=*buffer++;
        size -=sizeof(USHORT);
    }
    if(size)
        cksum += *(UCHAR*)buffer;

    cksum = (cksum >> 16) + (cksum & 0xffff);
    cksum += (cksum >>16);
    return (USHORT)(~cksum);
}

I too struggled to find c++/c code that computes it, until I found How to Calculate IP/TCP/UDP Checksum–Part 2 Implementation – roman10 , and it worked!我也很难找到计算它的 c++/c 代码,直到我找到了如何计算 IP/TCP/UDP 校验和 – 第 2 部分实现 – roman10 ,并且它奏效了! Tested it with Wireshark's validation.使用 Wireshark 的验证对其进行了测试。

UPDATE更新

Link broke meanwhile, recovered it and put it as a gist in my account - How to Calculate IP/TCP/UDP Checksum链接同时中断,将其恢复并将其作为要点放入我的帐户 - 如何计算 IP/TCP/UDP 校验和

I see a couple of things:我看到了几件事:

  • You are not accounting for odd length data by padding with zero.您没有通过用零填充来计算奇数长度数据。
  • You will need to account for network byte order when reading each word from the packet.从数据包中读取每个字时,您需要考虑网络字节顺序。
  • Your use of htonl(0x0000ffff) seems suspicious.您对htonl(0x0000ffff)似乎很可疑。 Why are you converting a constant to network byte order to combine it with data in host byte order?为什么要将常量转换为网络字节顺序以将其与主机字节顺序中的数据组合?

RFC 793 says "If a segment contains an odd number of header and text octets to be checksummed, the last octet is padded on the right with zeros to form a 16 bit word for checksum purposes." RFC 793说“如果一个段包含奇数个要校验和的报头和文本八位字节,最后一个八位字节在右边用零填充,以形成一个 16 位字,用于校验和目的。” Your code above does not handle that case.您上面的代码不处理这种情况。 I think the loop conditional should be i > 1 and then check for i == 1 outside the loop and do the special handling for the last octet.我认为循环条件应该是 i > 1 然后在循环外检查 i == 1 并对最后一个八位字节进行特殊处理。

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