开发者

Question about INADDR_ANY

The constant INADDR_ANY is the so-called IPv4 wildcard address. The wildcard IP address is useful for applications that bind Internet domain sockets on multihomed hosts. If an application on a multihomed host binds a socket to just one of its host’s IP addresses, then that socket can receive only UDP datagrams or TCP connection requests sent to that IP address. However, we normally want an application on a multihomed host to be able to receive datagrams or connection requests that specify any of the host’s IP addresses, and binding the socket to the wildcard IP address makes this possible.

struct sockaddr_in server_address;
int server_sockfd = socket(AF_INET, SOCK_STREAM, 0);
memset(&server_address, 0, sizeof(struct sockaddr_in));
server_address.sin_family = AF_INET;
server_address.sin_addr.s_addr = htonl(INADDR_ANY); // here is my quesion
server_address.sin_port = htons(9734);

bind(server_sockfd, (struct sockaddr*)&server_address, sizeof(server_address));

Question>

If we bind the socket to a specific IP address, then the socket can only receive UPD/TCP requests sent sent to that IP address.

As I show in the above code, now the socket server_sockfd is bound with INADDR_ANY. I just feel confused here b/c if the socket can receive any request on the internet, how it can still work well. There are tons of requests of UDP/TCP on internet, if the socket responses to everybody, , how can it still work?

// updated code for client side //

int
main(int argc, char *argv[])
{
    struct sockaddr_in6 svaddr;
    int sfd, j;
    size_t msgLen;
    ssize_t numBytes;
    char resp[BUF_SIZE];

    if (argc < 3 || strcmp(argv[1], "--help") == 0)
        usageErr("%s host-address msg...\n", argv[0]);

    /* Create a datagram socket; send to an address in the IPv6 somain */

    sfd = socket(AF_INET6, SOCK_DGRAM, 0);      /* Create client socket */
    if (sfd == -1)
        errExit("socket");

    memset(&svaddr, 0, sizeof(struct sockaddr_in6));
    svaddr.sin6_family = AF_INET6;
    svaddr.sin6_port = htons(PORT_NUM);
    if (inet_pton(AF_INET6, argv[1], &svaddr.sin6_addr) <= 0)
        fatal("inet_pton failed for address '%s'", argv[1]);

    /* Send messages to server; echo responses on stdout */

    for (j = 2; j < argc; j++) {
        msgLen = strlen(argv[j]);
        if (sendto(sfd, argv[j], msgLen, 0, (struct sockaddr *) &svaddr,
                    s开发者_如何学运维izeof(struct sockaddr_in6)) != msgLen)
            fatal("sendto");

        numBytes = recvfrom(sfd, resp, BUF_SIZE, 0, NULL, NULL);
        if (numBytes == -1)
            errExit("recvfrom");

        printf("Response %d: %.*s\n", j - 1, (int) numBytes, resp);
    }

    exit(EXIT_SUCCESS);
}

// updated for server side code

int
main(int argc, char *argv[])
{
    struct sockaddr_in6 svaddr, claddr;
    int sfd, j;
    ssize_t numBytes;
    socklen_t len;
    char buf[BUF_SIZE];
    char claddrStr[INET6_ADDRSTRLEN];

    /* Create a datagram socket bound to an address in the IPv6 somain */

    sfd = socket(AF_INET6, SOCK_DGRAM, 0);
    if (sfd == -1)
        errExit("socket");

    memset(&svaddr, 0, sizeof(struct sockaddr_in6));
    svaddr.sin6_family = AF_INET6;
    svaddr.sin6_addr = in6addr_any;                     /* Wildcard address */
    svaddr.sin6_port = htons(PORT_NUM);

    if (bind(sfd, (struct sockaddr *) &svaddr,
                sizeof(struct sockaddr_in6)) == -1)
        errExit("bind");

    /* Receive messages, convert to uppercase, and return to client */

    for (;;) {
        len = sizeof(struct sockaddr_in6);
        numBytes = recvfrom(sfd, buf, BUF_SIZE, 0,
                            (struct sockaddr *) &claddr, &len);
        if (numBytes == -1)
            errExit("recvfrom");

        /* Display address of client that sent the message */

        if (inet_ntop(AF_INET6, &claddr.sin6_addr, claddrStr,
                    INET6_ADDRSTRLEN) == NULL)
            printf("Couldn't convert client address to string\n");
        else
            printf("Server received %ld bytes from (%s, %u)\n",
                    (long) numBytes, claddrStr, ntohs(claddr.sin6_port));

        for (j = 0; j < numBytes; j++)
            buf[j] = toupper((unsigned char) buf[j]);

        if (sendto(sfd, buf, numBytes, 0, (struct sockaddr *) &claddr, len) !=
                numBytes)
            fatal("sendto");
    }
}

// updated for how to run this server/client programs.

$ ./server_program &
[1] 31047
$ ./client_program ::1 ciao // Send to server on local host
Server received 4 bytes from (::1, 32770)
Response 1: CIAO


It doesn't get requests for every IP address on the internet(a), it gets requests for every IP address that it services. For example, it may have multiple NICs, each with a separate IP address or it may have a single NIC capable of managing multiple IP addresses (it may even have multiple NICs, each capable of handling multiple IP addresses.

The key snippet to look at is:

... we normally want an application on a multi-homed host to be able to receive datagrams or connection requests that specify any of the host’s IP addresses (my italics).

In other words, you may have a multi-homed set-up where your machine services 10.0.0.15 and 10.0.0.16. Using INADDR_ANY will allow you to pick up traffic for both those addresses, without picking up requests for 10.0.0.17 which may be the machine on the other end of the bench (or other side of the planet).

The following table, with the top row being request destinations and the left column being the address you're listening on, shows whether you'll be given a request (Y) or not (N):

Request to>  10.0.0.15  10.0.0.16  10.0.0.17
Bind to:    *-------------------------------
10.0.0.15   |    Y          N          N
10.0.0.16   |    N          Y          N
INADDR_ANY  |    Y          Y          N

(a) It doesn't even see the vast majority of requests on the net. The vast majority don't even make it to your nearest router (or probably even your ISP). Even those that do make it to your nearest router, your particular machine might not see if they're destined for another machine on the local segment (promiscuous mode notwithstanding).

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜