Question

i 阅读了《打字法》。 i米混淆

    *(u_char *)(&u) = *(u_char *)w ;

在这项职能中,我认为“u”的价值为0,因此重新分派。

/*
 *                      I N _ C K S U M
 *
 * Checksum routine for Internet Protocol family headers (C Version)
 *
 */
in_cksum(addr, len)
u_short *addr;
int len;
{
        register int nleft = len;
        register u_short *w = addr;
        register u_short answer;
        register int sum = 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_short u = 0;

                *(u_char *)(&u) = *(u_char *)w ;
                sum += u;
        }

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

Answer 1

&u

http://code>u。

(u_char *)(&u)

au_char/code> pointer to u。



*(u_char *)(&u)


参照点,

*(u_char *)(&u) = *(u_char *)w ;


http://www.un.org/Depts/DGACM/index_french.htm 第二步是零。

(如本法典作者所做的那样,假设short为16比喻。) 根据C标准,这不一定是真实的。

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