getaddrinfo和IPv6

Question

我试图了解getaddrinfo函数返回的内容:

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

int main (int argc, char *argv[])
{


struct addrinfo *res = 0 ;

  getaddrinfo("localhost", NULL ,NULL,&res);
  printf("ai_flags -> %i\n", res->ai_flags) ;
  printf("ai_family -> %i\n", res->ai_family) ;
  printf("ai_socktype -> %i\n", res->ai_socktype) ;
  printf("ai_protocol -> %i\n", res->ai_protocol) ;
  printf("ai_addrlen -> %i\n", res->ai_addrlen) ;
  struct sockaddr_in* saddr = (struct sockaddr_in*)res->ai_addr;
  printf("ai_addr hostname ->  %s\n", inet_ntoa(saddr->sin_addr));

  freeaddrinfo(res);

  return 0 ;
}

结果:

ai_flags -> 40
ai_family -> 2
ai_socktype -> 1
ai_protocol -> 6
ai_addrlen -> 16
ai_addr hostname ->  127.0.0.1

在/ etc/hosts中,我有:

127.0.0.1 localhost    
::1     localhost

Getaddrinfo只返回127.0.0.1而不是:: 1？我不明白为什么？

第二个问题是我在哪里可以找到这些整数的含义(40,2,1,6等)？我读过这个男人,但那里什么都没有.

我还想知道是否可以提供IPv6地址(例如:: 1)并且该函数返回名称:localhost？

非常感谢 !!

Answer 1

@jwodder和@onteria_很好地覆盖了IPv6部分,所以我只是解决数字部分:

ai_flags -> 40

可能这将是以下两个的总和/usr/include/netdb.h:

# define AI_V4MAPPED    0x0008  /* IPv4 mapped addresses are acceptable.  */
# define AI_ADDRCONFIG  0x0020  /* Use configuration of this host to choose

这是协议族,inet,inet6,apx,unix等:

ai_family -> 2

bits/socket.h:78:#define    PF_INET     2   /* IP protocol family.  */
bits/socket.h:119:#define   AF_INET     PF_INET

这是socket类型,流,dgram,包,rdm,seqpacket:

ai_socktype -> 1

bits/socket.h:42:  SOCK_STREAM = 1,     /* Sequenced, reliable, connection-based

更高级别的协议,TCP,UDP,TCP6,UDP6,UDPlite,ospf,icmp等:

ai_protocol -> 6

很有趣,在/etc/protocols:

tcp 6   TCP     # transmission control protocol

的大小struct sockaddr.(根据地址系列不同!呃.)

ai_addrlen -> 16

这是因为你要回来了struct sockaddr_in,看看linux/in.h:

#define __SOCK_SIZE__   16      /* sizeof(struct sockaddr)  */
struct sockaddr_in {
  sa_family_t       sin_family; /* Address family       */
  __be16        sin_port;   /* Port number          */
  struct in_addr    sin_addr;   /* Internet address     */

  /* Pad to size of `struct sockaddr'. */
  unsigned char     __pad[__SOCK_SIZE__ - sizeof(short int) -
            sizeof(unsigned short int) - sizeof(struct in_addr)];
};

最后一个,来自/etc/hosts:)

ai_addr hostname ->  127.0.0.1

Answer 2

res还包含一个字段struct addrinfo *ai_next;,它是指向找到的其他条目的指针getaddrinfo,如果没有其他条目,则为NULL.如果您检查res->ai_next,您应该找到IPv6条目.

对于a中的整数字段struct addrinfo,它们对应于具有实现定义值的预定义常量,并且整数值本身不是普遍感兴趣的.如果你想知道一个给定的字段表示,比较其对可分配给该字段的常数(SOCK_STREAM,SOCK_DGRAM,等了ai_socktype; IPPROTO_TCP,IPPROTO_UDP等了ai_protocol;等等),或者为ai_flags,测试对应于每个位预定义常量(例如if (res->ai_flags & AI_NUMERICHOST) {printf("ai_flags has AI_NUMERICHOST\n"); }).

Answer 3

extern struct sockaddr_in6 create_socket6(int port, const char * address) {

    struct addrinfo hints, *res, *resalloc;
    struct sockaddr_in6 input_socket6;
    int errcode;

    /* 0 out our structs to be on the safe side */
    memset (&hints, 0, sizeof (hints));
    memset (&input_socket6, 0, sizeof(struct sockaddr_in6));

    /* We only care about IPV6 results */
    hints.ai_family = AF_INET6;
    hints.ai_socktype = SOCK_STREAM;
    hints.ai_flags = AI_DEFAULT;

    errcode = getaddrinfo (address, NULL, &hints, &res);
    if (errcode != 0)
    {
     perror ("[ERROR] getaddrinfo ");
     return input_socket6;
    }

    resalloc = res;

    while (res)
    {
        /* Check to make sure we have a valid AF_INET6 address */
        if(res->ai_family == AF_INET6) {
                /* Use memcpy since we're going to free the res variable later */
                        memcpy (&input_socket6, res->ai_addr, res->ai_addrlen);

                       /* Here we convert the port to network byte order */
                        input_socket6.sin6_port = htons (port);
                        input_socket6.sin6_family = AF_INET6;
               break;
        }

        res = res->ai_next;
    }

    freeaddrinfo(resalloc);

    return input_socket6;
}

这是一些解释它的代码.基本上除非你给getaddrinfo一些提示告诉它只能使用IPV6,它也会给出IPV4结果.这就是为什么你必须如图所示循环结果.