从 URL 获取 IP (C++)

Question

如何从具有顶级域名的域名中获取 IP 地址？ 在本例中获取 google.com 的 IP。 如果可能的话，在 IPv6 中以正确的格式。

这是我迄今为止尝试过的：

#include <netdb.h>

using namespace std;

int main()
{
    struct hostent* myHostent;
    myHostent = gethostbyname("google.com");
    cout<<myHostent <<"\n";
    //output is a hex code
    cout<<myHostent->h_name<<"\n";
    //output is google.com
    cout<<myHostent->h_aliases;
    //output is a hex code  
}

Answer 1

域的 IP 地址（是复数，可以超过 1 个）在hostent::h_addr_list字段中，而不在hostent::h_aliases字段中，例如：

int main()
{
    hostent* myHostent = gethostbyname("google.com");
    if (!myHostent)
    {
        cerr << "gethostbyname() failed" << "\n";
    }
    else
    {
        cout << myHostent->h_name << "\n";

        char ip[INET6_ADDRSTRLEN];
        for (unsigned int i = 0; myHostent->h_addr_list[i] != NULL; ++i)
        {
            cout << inet_ntop(myHostent->h_addrtype, myHostent->h_addr_list[i], ip, sizeof(ip)) << "\n";
        }
    }

    return 0;
}

也就是说，不推荐使用gethostbyname() ，请改用getaddrinfo() ，例如：

void* getSinAddr(addrinfo *addr)
{
    switch (addr->ai_family)
    {
        case AF_INET:
            return &(reinterpret_cast<sockaddr_in*>(addr->ai_addr)->sin_addr);

        case AF_INET6:
            return &(reinterpret_cast<sockaddr_in6*>(addr->ai_addr)->sin6_addr);
    }

    return NULL;
}

int main()
{
    addrinfo hints = {};
    hints.ai_flags = AI_CANONNAME;
    hints.ai_family = AF_UNSPEC;
    hints.ai_socktype = SOCK_STREAM;
    hints.ai_protocol = IPPROTO_TCP;

    addrinfo *res;

    int ret = getaddrinfo("google.com", NULL, &hints, &res);
    if (ret != 0)
    {
        cerr << "getaddrinfo() failed: " << gai_strerror(ret) << "\n";
    }
    else
    {
        cout << res->ai_canonname << "\n";

        char ip[INET6_ADDRSTRLEN];
        for(addrinfo addr = res; addr; addr = addr->ai_next)
        {
            cout << inet_ntop(addr->ai_family, getSinAddr(addr), ip, sizeof(ip)) << "\n";
        }

        freeaddrinfo(res);
    }

    return 0;
}