12.6 如何手动计算ip_sum

https://scz.617.cn/network/200611071801.txt

A: W. Richard Stevens

--------------------------------------------------------------------------
static unsigned short int in_cksum
(
    unsigned short int *addr,
    unsigned int        len
)
{
    unsigned int        nleft   = len;
    unsigned int        sum     = 0;
    unsigned short int *w       = addr;
    unsigned short int  answer  = 0;

    /*
     * Our algorithm is simple, using a 32 bit accumulator (sum), we add
     * sequential 16 bit words to it, and at the end, fold back all the
     * carry bits from the top 16 bits into the lower 16 bits.
     */
    while ( nleft > 1 )
    {
        sum    += *w++;
        nleft  -= 2;
    }
    /*
     * 4mop up an odd byte, if necessary
     */
    if ( 1 == nleft )
    {
        *( unsigned char * )( &answer )     = *( unsigned char * )w ;
        sum                                += answer;
    }
    /*
     * 4add back carry outs from top 16 bits to low 16 bits
     */
    /*
     * add hi 16 to low 16
     */
    sum     = ( sum >> 16 ) + ( sum & 0xffff );
    /*
     * add carry
     */
    sum    += ( sum >> 16 );
    /*
     * truncate to 16 bits
     */
    answer  = ~sum;
    return( answer );
}  /* end of in_cksum */
--------------------------------------------------------------------------

A: xjojox@netexpert 2006-11-07 18:01

以如下IP首部为例:

45 00 00 34 1F CF 40 00 40 06 3C A9 DA F6 BC FD DA 1E 6C 39
                              ~~~~~
做如下运算:

    45 00
    00 34
    1F CF
    40 00
    40 06
            3C A9(这个字段不要参与运算)
    DA F6
    BC FD
    DA 1E
  + 6C 39
---------
 03 C3 53

将进位加回:

    C3 53
  +    03
---------
    C3 56

求反码得校验和:

    3C A9

D: 2009-11-05 16:28 刘昶

将进位加回:

    C3 53
  +    03
---------
    C3 56

这一步可能仍有进位，还需要将这个进位加回。最极端的情况是:

    FF FF
  + FF FF
---------
 01 FF FE

此时的进位最多是1，将32-bits表示的结果右移16位之后加回去即可，不会再次产生
进位:

    FF FE
  +    01
---------
    FF FF

这个算法基于一个假设，最开始的短整型累加和不会超过0xFFFFFFFF。就IP报文来说
这是可以保证的。即使每一个短整型都是0xFFFF也无所谓，因为ip_len本身是短整型，
限制了IP报文的总字节数，0xFFFF * ( 0xFFFF / 2 ) + 0xFF = 0x7FFE8100。下面
是一种更接近真实的极端情况:

      FF FF
      FF FF
            \
      ...    (0x7FFF个0xFFFF累加，最后再加一个0xFF)
            /
      FF FF
      FF FF
    +    FF
-----------
7F FE 81 00

将进位加回:

    81 00
  + 7F FE
---------
 01 00 FE

再次将进位加回:

    00 FE
  +    01
---------
    00 FF

求反码得校验和:

    FF 00