c - 为什么我不能在 ai_socktype 中使用 SOCK_RAW？

Question

我正在尝试实现一个 IPv6 UDP 客户端-服务器应用程序，它将让我获得逐跳扩展 header 的详细信息。因此，为此我想我必须使用 SOCK_RAW 来访问 IPv6 header 信息。

现在在我的客户端代码中，我有以下

   #include <sys/types.h>
   #include <sys/socket.h>
   #include <netdb.h>
   #include <stdio.h>
   #include <stdlib.h>
   #include <unistd.h>
   #include <string.h>

   #define BUF_SIZE 500

   int
   main(int argc, char *argv[])
   {
       struct addrinfo hints;
       struct addrinfo *result, *rp;
       int sfd, s, j;
       size_t len;
       ssize_t nread;
       char buf[BUF_SIZE];

       if (argc < 3) {
           fprintf(stderr, "Usage: %s host port msg...\n", argv[0]);
           exit(EXIT_FAILURE);
       }

       /* Obtain address(es) matching host/port */

       memset(&hints, 0, sizeof(struct addrinfo));
       hints.ai_family = AF_INET6;    /* Allow IPv4 or IPv6 */
       hints.ai_socktype = SOCK_RAW; /* Datagram socket */
       hints.ai_flags = 0;
       hints.ai_protocol = 0;          /* Any protocol */

       s = getaddrinfo(argv[1], argv[2], &hints, &result);
       if (s != 0) {
           fprintf(stderr, "getaddrinfo: %s\n", gai_strerror(s));
           exit(EXIT_FAILURE);
       }

       /* getaddrinfo() returns a list of address structures.
          Try each address until we successfully connect(2).
          If socket(2) (or connect(2)) fails, we (close the socket
          and) try the next address. */

       for (rp = result; rp != NULL; rp = rp->ai_next) {
           sfd = socket(rp->ai_family, rp->ai_socktype,
                        rp->ai_protocol);
           if (sfd == -1)
               continue;

           if (connect(sfd, rp->ai_addr, rp->ai_addrlen) != -1)
               break;                  /* Success */

           close(sfd);
       }

       if (rp == NULL) {               /* No address succeeded */
           fprintf(stderr, "Could not connect\n");
           exit(EXIT_FAILURE);
       }

       freeaddrinfo(result);           /* No longer needed */

       /* Send remaining command-line arguments as separate
          datagrams, and read responses from server */

       for (j = 3; j < argc; j++) {
           len = strlen(argv[j]) + 1;
                   /* +1 for terminating null byte */

           if (len + 1 > BUF_SIZE) {
               fprintf(stderr,
                       "Ignoring long message in argument %d\n", j);
               continue;
           }

           if (write(sfd, argv[j], len) != len) {
               fprintf(stderr, "partial/failed write\n");
               exit(EXIT_FAILURE);
           }

           nread = read(sfd, buf, BUF_SIZE);
           if (nread == -1) {
               perror("read");
               exit(EXIT_FAILURE);
           }

           printf("Received %ld bytes: %s\n", (long) nread, buf);
       }

       exit(EXIT_SUCCESS);
   }

我知道在使用 SOCK_RAW 时我没有为数据包编写 header 。我只是想试运行一下，看看我得到的错误是什么。它在 getaddrinfo() 中失败并给出以下错误，

getaddrinfo(): ai_socktype 不支持 servname

我担心的是，它不应该在这里失败，而应该是在它向套接字发送一些数据或者可能在创建套接字时失败。

这可能是什么原因...？

对于在 IPv6 中构建数据包时的 SOCK_RAW，我是否也需要处理校验和或由内核完成。从我目前所读的内容来看，必须为 ICMPv6 处理校验和计算。

请让我知道我是否做对了事情以及我是否正朝着正确的方向前进。

Answer 1

我猜这会失败，因为您用两个参数调用它。

来自 getaddrinfo 的手册页:

The requested service is not available for the requested socket type. [...] the error could occur if service was not NULL, and hints.ai_socktype was SOCK_RAW (a socket type that does not support the concept of services).

只给 NULL 而不是 argv[2]