任何人都可以指出你不能同时在套接字上发送和接收的原因吗?
据我所知,有2个流可以推送,1个可以拉如果您尝试同时发送/接收,则会出现wasealready错误
套接字抛出wasealready错误的原因是什么(10035) 它与接收方发回的ack窗口有什么关系吗? 好像要为窗户打开线?
答案 0 :(得分:2)
首先,我认为您对错误代码感到困惑。以下是一些可能相关的代码:
connect()
。阻止您一次执行两个阻塞操作的主要因素是WSAEINPROGRESS
错误 - Winsock 1.1 API中的限制只允许您一次执行一个阻塞操作。您可以通过使用非阻塞重叠调用和Winsock 2.0 API来模拟阻塞调用来解决此问题。例如:
SOCKET sock = WSASocket(AF_INET, SOCK_STREAM, IPPROTO_TCP, NULL, 0, WSA_FLAG_OVERLAPPED);
WSAEVENT hEvent = WSACreateEvent();
WSAEventSelect(sock, hEvent, FD_CONNECT); // register for connect notification
long iMode=1;
ioctlsocket(sock,FIONBIO,&iMode); // set non-blocking mode
WSAConnect(sock, address, sizeof(address), NULL, NULL, NULL, NULL);
WSAWaitForMultipleEvents(1, &events, FALSE, INFINITE, FALSE);
WSANETWORKEVENTS connect_result;
WSAEnumNetworkEvents(sock, hEvent, &connect_result);
if (connect_result.iErrorCode[FD_CONNECT] != 0) {
// connect failed, do something about it
}
WSACloseEvent(hEvent);
iMode = 0;
ioctlsocket(sock,FIONBIO,&iMode); // clear non-blocking mode; it won't play well with overlapped IO below
// Start an overlapped send
char *my_data = "Hello, world!";
WSAOVERLAPPED send_overlapped;
WSABUF send_buf;
send_buf.len = strlen(my_data);
send_buf.buf = my_data;
send_overlapped.hEvent = WSACreateEvent();
DWORD bytesSent;
WSASend(sock, &send_buf, 1, &bytesSent, 0, &send_overlapped, NULL);
if (WSAGetLastError() != WSA_IO_PENDING) {
// something went wrong, handle the error...
}
// Start an overlapped receive
char rdata[256];
WSAOVERLAPPED recv_overlapped;
WSABUF recv_buf;
recv_buf.len = sizeof(rdata);
recv_buf.buf = rdata;
recv_overlapped.hEvent = WSACreateEvent();
DWORD bytesRecvd;
WSARecv(sock, &recv_buf, 1, &bytesRecvd, NULL, &recv_overlapped, NULL);
if (WSAGetLastError() != WSA_IO_PENDING) {
// something went wrong, handle the error...
}
// Now wait for both to finish
WSAEVENT events[2];
events[0] = recv_buf.hEvent;
events[1] = send_buf.hEvent;
WSAWaitForMultipleEvents(2, events, TRUE, INFINITE, TRUE);
这种方法有点复杂,但非常灵活 - 一旦你发起重叠的I / O作业,同一个线程可以继续做其他工作(只要你保留缓冲区和未分配的内容! )。需要注意的一件事是,如果你在GUI线程上执行此操作,则需要更改消息循环 - 为了完成重叠事件,发出它们的线程需要使用例如{执行警报等待{1}}。在上面的示例中,我使用MsgWaitForMultipleEvents
来执行此警报等待。此外,发出重叠I / O请求的线程在请求完成之前不得终止。
有关重叠I / O的更多常规信息,请参阅此MSDN页面:http://msdn.microsoft.com/en-us/library/ms686358(v=vs.85).aspx