c++ - 如何从 IPv4 映射的 IPv6 地址解析 IPv4 地址？

Question

如何从 IPv4 映射的 IPv6 地址获取 IPv4 地址？

例如，我有一个 IP 地址 ::FFFF:129.144.52.38。从中，我需要提取 129.144.52.38。是否有任何用于此目的的 API？

我可以使用以下函数识别 IPv6 或 IPv4 地址族

int getaddrfamily(const char *addr)
{
    struct addrinfo hint, *info =0;
    memset(&hint, 0, sizeof(hint));
    hint.ai_family = AF_UNSPEC;
    // Uncomment this to disable DNS lookup
    //hint.ai_flags = AI_NUMERICHOST;
    int ret = getaddrinfo(addr, 0, &hint, &info);
    if (ret)
        return -1;
    int result = info->ai_family;
    freeaddrinfo(info);
    return result;
}

如果我给出一个 IPv4 映射的 IPv6 地址，那么如何确定它是否是一个映射地址？是否有任何套接字 API 可以从映射的 IPv6 地址中提取 IPv4？

Answer 1

尝试这样的事情:

#ifndef IN6_IS_ADDR_V4MAPPED
#define IN6_IS_ADDR_V4MAPPED(a) \
       ((((a)->s6_words[0]) == 0) && \
        (((a)->s6_words[1]) == 0) && \
        (((a)->s6_words[2]) == 0) && \
        (((a)->s6_words[3]) == 0) && \
        (((a)->s6_words[4]) == 0) && \
        (((a)->s6_words[5]) == 0xFFFF))
#endif

unsigned long getIPv4addr(const char *addr)
{
    struct addrinfo hint, *info = 0;
    unsigned long result = INADDR_NONE;
    memset(&hint, 0, sizeof(hint));
    hint.ai_family = AF_UNSPEC;
    // Uncomment this to disable DNS lookup
    //hint.ai_flags = AI_NUMERICHOST;
    if (getaddrinfo(addr, 0, &hint, &info) == 0)
    {
        switch (info->ai_family)
        {
            case AF_INET:
            {
                struct sockaddr_in *addr = (struct sockaddr_in*)(info->ai_addr);
                result = addr->sin_addr.s_addr;
                break;
            }

            case AF_INET6:
            {
                struct sockaddr_in6 *addr = (struct sockaddr_in6*)(info->ai_addr);
                if (IN6_IS_ADDR_V4MAPPED(&addr->sin6_addr))
                    result = ((in_addr*)(addr->sin6_addr.s6_addr+12))->s_addr;
                break;
            }
        }
        freeaddrinfo(info);
    }
    return result;
}