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。由此，我需要提取129.144.52.38。是否有用于此目的的API？

我可以使用以下功能识别IPv6或IPv4地址系列

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;
}

如果我提供IPv4映射的IPv6地址，那么如何识别它是否是映射的地址？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_word[2]) == 0) && \
        (((a)->s6_word[3]) == 0) && \
        (((a)->s6_word[4]) == 0) && \
        (((a)->s6_word[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;
}

Answer 2

您只需从IPv6地址中提取最后四个字节，将它们合并为一个32位数字，然后就可以获得IPv4地址。

当然，您需要先检查它是否真的是IPv4映射地址。

如何从IPv4映射的IPv6地址解析IPv4地址？

2 个答案: