Checksum ip Packet

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Guys would like to understand the 4 last lines , why do all these bitswise? Code complete.

Follows the code:

static int in_cksum(u_short *addr, int len)
{
    register int nleft = len;
    register u_short *w = addr;
    register int sum = 0;
    u_short 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;
    }

    /* mop up an odd byte, if necessary */
    if (nleft == 1) {
        *(u_char *)(&answer) = *(u_char *)w ;
        sum += answer;
    }


    sum = (sum >> 16) + (sum & 0xffff); /* add hi 16 to low 16 */
    sum += (sum >> 16);         /* add carry */
    answer = ~sum;              /* truncate to 16 bits */
    return(answer);
}
c

  • I think it would be better to have edited the previous one instead of asking an almost equal question. As this got a little better, it might be better to remove that. http://answall.com/questions/111161/

1 answer

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After the initial calculations completed, the variable sum (32 bits) will have any value.

The first expression (sum = (sum >> 16) + (sum & 0xffff);) isolates the two 16-bit halves of this value and sums them, putting the result in the original variable

bits    31 30 ... 17 16 // sum >> 16
bits  + 15 14 ... 01 00 // sum & 0xffff
      -----------------
bits xx 15 14 ... 01 00 // xx é o carry

Note that the sum of these two halves will have a maximum of 17 significant bits.

The second expression (sum += (sum >> 16);) makes (more or less) the same as the first. The difference is that this expression does not care about what remains in high bits.

bits    31 30 ... 17 16 15 14 ... 01 00 // sum
bits  +                 31 30 ... 17 16 // sum >> 16
      ---------------------------------
bits xx xx ... xx xx xx 15 14 ... 01 00 // xx é o carry nos high bits

Finally, the third expression (answer = ~sum;) change the calculated bits and isolate the 16 low bits because it puts the value in a type variable u_short.

answer = ~sum;
// bits de ~sum:   31 30 ... 17 16 15 14 ... 01 00
// bits de answer:                 15 14 ... 01 00

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