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。从这里，我需要提取129.144.52.38。有没有用于此目的的API？

我可以使用下面的函数

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; 
} 

识别IPv6或IPv4地址族如果我给一个IPv4映射的IPv6地址，这又如何可能识别，如果它是一个映射ADRESS？是否有任何套接字API从映射的IPv6地址中提取IPv4？

Answer 1

尝试这样：

#ifndef IN6_IS_ADDR_V4MAPPED 
#define IN6_IS_ADDR_V4MAPPED(a) \ 
     ((((a)->s6_words[0]) == 0) && \ 
     (((a)->s6_words[1]) == 0) && \ 
     (((a)->s6_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，并且您具有IPv4地址。