如何从IPv4映射的IPv6地址中解析IPv4地址

How to resolve IPv4 address from IPv4 mapped IPv6 address?

本文关键字：IPv4 地址 IPv6 映射      更新时间：2023-10-16

如何从IPv4映射的IPv6地址中获取IPv4地址?

例如，我有一个IP地址::FFFF:129.144.52.38。从这里，我需要提取129.144.52.38。有这样的API吗?

我可以通过使用以下函数来识别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;
}

如果我给一个IPv4映射的IPv6地址，那么它是如何可能识别，如果它是一个映射的地址?是否有任何套接字API从映射的IPv6地址提取IPv4 ?

试试这样:

#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;
}

您只需从IPv6地址中提取最后四个字节，将它们组合成单个32位数字，就可以得到您的IPv4地址。

当然你需要先检查它是否真的是ipv4映射地址

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