我正试图构建一个工具,根据域名收集有关IP地址的信息。我一直在使用inet_ntop来实现这一目标,并且通常在大多数领域都能很好地工作。当我在google.com或amazon.com等网站上尝试时,相同的IP地址似乎会被反复打印。
输出示例:主机: google.com IPv4: 54.239.17.6 .
我想知道是否有人经历过这种情况,是我做错了什么,还是这与网站的配置方式有关。
error = getaddrinfo(domain, "80", &host_info, &host_list);
// if error exists exit gracefully
if(error != 0)
{
print_conn_err(error);
return EXIT_FAILURE;
}
// print the domain string array
printf("Host: %s\n", domain);
// print IP addresses relating to domain name
printf("IP Address(s):\n");
for(result = host_list; result != NULL; result = host_list->ai_next)
{
void * address;
char * ip_version;
// if result is IPv4 address
if(result->ai_family == AF_INET)
{
struct sockaddr_in * ipv4 = (struct sockaddr_in *)result->ai_addr;
address = &(ipv4->sin_addr);
ip_version = "IPv4";
}
else if(result->ai_family == AF_INET6)
{
struct sockaddr_in6 * ipv6 = (struct sockaddr_in6 *)result->ai_addr;
address = &(ipv6->sin6_addr);
ip_version = "IPv6";
}
else
{
printf("%s", "Not a valid IP addresses \n.");
return EXIT_FAILURE;
}
// convert the IP to a string and print it:
inet_ntop(result->ai_family, address, ip_address, sizeof ip_address);
printf(" %s: %s\n", ip_version, ip_address);
}
freeaddrinfo(result);我想到的一个想法是用以前的IP检查当前的IP并中断循环。不过,这似乎是一个相当麻烦的解决方案。
发布于 2015-12-30 19:11:13
您的代码中有一个bug导致了这种情况。
for(result = host_list; result != NULL; result = host_list->ai_next)需要的是
for(result = host_list; result != NULL; result = result->ai_next)您的freeaddrinfo(result)也必须更改为freeaddrinfo(host_list)
https://stackoverflow.com/questions/34535611
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