[英]how do you cast sockaddr structure to a sockaddr_in - C++ networking sockets ubuntu UDP
我想獲取客戶端地址,但我不確定如何將sockaddr結構轉換為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);
提前致謝! :)
我發現問題讓我走到了這一步: 從sockaddr結構中獲取IPV4地址
很抱歉為了避免混淆,這是我真正的實現,其中“ci”是存儲指針(如sockaddr_in)的對象。
/* receive message */
n = recvfrom(*(ci->getSd()), msg, MAX_MSG, 0,(struct sockaddr *) ci->getCliAddr(),ci->getCliLen());
char *ip = inet_ntoa(ci->getCliAddr().sin_addr);
我會收到以下錯誤:
udpserv.cpp:166: error: request for member ‘sin_addr’ in ‘ci->clientInfo::getCliAddr()’, which is of non-class type ‘sockaddr_in*’
我想指出,如果這實際上是C ++,那么慣用的方法是:
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();
}
此示例代碼處理IPv4和IPv6地址,並且還被認為比任何建議的實現都更加C ++慣用。
它實際上非常簡單!
struct sockaddr *sa = ...;
if (sa->sa_family == AF_INET)
{
struct sockaddr_in *sin = (struct sockaddr_in *) sa;
ip = inet_ntoa(sin->sin_addr);
}
我認為這將為你編譯得很好,並做你想要的。
struct sockaddr_in cliAddr={}, servAddr={};
socklen_t cliAddrLength = sizeof(cliAddr);
n = recvfrom(sd, msg, MAX_MSG, 0,(struct sockaddr *)&cliAddr, &cliAddrLength);
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