我在dualstack套接字上使用TCP accept()函数。接下来我尝试使用getnameinfo()打印IPV4映射的Ipv6客户端地址。对于第一个连接到accept(),getnameinfo()正确返回IPv4-Ipv6客户端源地址。但对于病房中的第二个客户端,它首先返回相同的IPv4映射的Ipv6地址。任何人都可以帮助我吗?
伪代码:
struct sockaddr_storage client_ip;
socklen_t sock_len = sizeof(client_ip);
char buffer[INET6_ADDRSTRLEN];
struct sockaddr_in sa4;
static char *pclient_ipv4_addr = NULL;
while(1)
{
fd = accept(master_fd, (struct sockaddr*)&client_ip, &sock_len);
if (!getnameinfo((struct sockaddr*)&client_ip, sock_len, buffer,
sizeof(buffer),0,0,NI_NUMERICHOST))
{
pclient_ipv4_addr = buffer;
if (client_ip.ss_family==AF_INET6)
{
struct sockaddr_in6 *sa6=(struct sockaddr_in6*)&client_ip;
if (IN6_IS_ADDR_V4MAPPED(&sa6->sin6_addr))
{
memset(&sa4,0,sizeof(sa4));
sa4.sin_family=AF_INET;
sa4.sin_port=sa6->sin6_port;
memcpy(&sa4.sin_addr.s_addr, sa6->sin6_addr.s6_addr+12, 4);
memcpy(&client_ip, &sa4,sizeof(sa4));
sock_len=sizeof(sa4);
pclient_ipv4_addr = inet_ntoa(sa4.sin_addr);
printf("IPV4 add is %s\n", pclient_ipv4_addr );
}
}
}
memset(buffer, 0 , INET6_ADDRSTRLEN);
}
例如:对于第一个客户端,pclient_ipv4_addr是“192.168.208.13”,那么对于下一个客户端也是 pclient_ipv4_addr正在打印相同的地址。不可重入getnameinfo(0任何问题?如何 能解决这个问题?
答案 0 :(得分:1)
尝试更像这样的事情:
struct sockaddr_storage client_ip;
socklen_t sock_len;
while(1)
{
sock_len = sizeof(client_ip);
fd = accept(master_fd, (struct sockaddr*)&client_ip, &sock_len);
if (fd == -1)
break;
if (client_ip.ss_family==AF_INET)
{
struct sockaddr_in *sa4 = (struct sockaddr_in*) &client_ip;
printf("IPv4 add is %s\n", inet_ntoa(sa4->sin_addr) );
}
else if (client_ip.ss_family==AF_INET6)
{
struct sockaddr_in6 *sa6 = (struct sockaddr_in6*) &client_ip;
if (IN6_IS_ADDR_V4MAPPED(&sa6->sin6_addr))
{
struct in_addr a4;
memcpy(&a4, sa6->sin6_addr.s6_addr+12, 4);
printf("IPv4 add is %s\n", inet_ntoa(a4) );
}
else
{
char buffer[INET6_ADDRSTRLEN];
if (getnameinfo((struct sockaddr*)&client_ip, sock_len, buffer, sizeof(buffer), 0, 0, NI_NUMERICHOST) == 0)
printf("IPv6 add is %s\n", buffer );
}
}
}