如何将sockaddr结构转换为sockaddr_in - C++网络套接字ubuntu UDP

Question

我想获取客户端地址,但我不确定如何将sockaddr结构转换为sockaddr_in？

struct sockaddr_in cliAddr, servAddr;

    n = recvfrom(sd, msg, MAX_MSG, 0,(struct sockaddr *) cliAddr,sizeof(cliAddr));

 //i tried this but it does not work
    struct sockaddr cliSockAddr = (struct sockaddr *) cliAddr; 
    char *ip = inet_ntoa(cliSockAddr.sin_addr);

提前致谢!:)

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我发现问题让我走到了这一步: 从sockaddr结构中获取IPV4地址

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很抱歉为了避免混淆,这是我真正的实现,其中"ci"是存储指针(如sockaddr_in)的对象.

    /* receive message */
    n = recvfrom(*(ci->getSd()), msg, MAX_MSG, 0,(struct sockaddr *) ci->getCliAddr(),ci->getCliLen());

    char *ip = inet_ntoa(ci->getCliAddr().sin_addr);

我会收到以下错误:

udpserv.cpp:166: error: request for member ‘sin_addr’ in ‘ci->clientInfo::getCliAddr()’, which is of non-class type ‘sockaddr_in*’

Answer 1

我想指出,如果这实际上是C++,那么惯用的方法是:

sockaddr *sa = ...; // struct not needed in C++
char ip[INET6_ADDRSTRLEN] = {0};

switch (sa->sa_family) {
  case AF_INET: {
    // use of reinterpret_cast preferred to C style cast
    sockaddr_in *sin = reinterpret_cast<sockaddr_in*>(sa);
    inet_ntop(AF_INET, &sin->sin_addr, ip, INET6_ADDRSTRLEN);
    break;
  }
  case AF_INET6: {
    sockaddr_in6 *sin = reinterpret_cast<sockaddr_in6*>(sa);
    // inet_ntoa should be considered deprecated
    inet_ntop(AF_INET6, &sin->sin6_addr, ip, INET6_ADDRSTRLEN);
    break;
  }
  default:
    abort();
}

此示例代码处理IPv4和IPv6地址,并且还被认为比任何建议的实现都更加C++惯用.

Answer 2

它实际上非常简单!

struct sockaddr *sa = ...;

if (sa->sa_family == AF_INET)
{
    struct sockaddr_in *sin = (struct sockaddr_in *) sa;
    ip = inet_ntoa(sin->sin_addr);
}