基础问题,connect函数什么时候返回?

[saiyaman5]

我的理解:

服务器socket listen 之后,
客户端connect ,但要在服务器accept 之后,客户端connect才会返回(阻塞情况下)

测试结果:
服务器只置listen状态,没有accept操作,客户端connect马上返回0(表示成功)


connect返回0,表示告诉客户端,连接已经成功,但似乎....?
当然,服务器与客户端都在同一台机器上操作,不知道有没有影响

[Wenxy1]

理论上是不可能的。
请贴上你的代码。

[saiyaman5]

服务器端

// Initialize Winsock
	WSADATA wsaData;
	WSAStartup(MAKEWORD(2,2), &wsaData);

	//-----------------------------------------
	// Create a listening socket bound to a local
	// IP address and the port specified
	ListenSocket = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
	u_short port = 6000;
	char* ip;
	sockaddr_in service;
	service.sin_family = AF_INET;
	service.sin_port = htons(port);
	/*hostent* thisHost;
	thisHost = gethostbyname("");
	ip = inet_ntoa (*(struct in_addr *)*thisHost->h_addr_list);

	service.sin_addr.s_addr = inet_addr(ip);*/

	service.sin_addr.s_addr = htonl(INADDR_ANY);

	//-----------------------------------------
	// Bind the listening socket to the local IP address
	// and port number
	bind(ListenSocket, (SOCKADDR *) &service, sizeof(SOCKADDR));

	//-----------------------------------------
	// Set the socket to listen for incoming
	// connection requests
	listen(ListenSocket, 1);
	printf("Listening...\n");
    Sleep(100000);    

客户端

	WSADATA wsaData;
	int iResult = WSAStartup(MAKEWORD(2,2), &wsaData);
	SOCKET ConnectSocket;
		ConnectSocket = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
		if (ConnectSocket == INVALID_SOCKET) 
		{
			return;
		}

		sockaddr_in clientService; 
		clientService.sin_family = AF_INET;
		clientService.sin_addr.s_addr = inet_addr( "127.0.0.1" );
		clientService.sin_port = htons( 6000 );

		int nRet = connect( ConnectSocket, (SOCKADDR*) &clientService, sizeof(clientService) );

		if ( nRet == SOCKET_ERROR) {
			DWORD dw = GetLastError();
			return;
		}

		printf("socket %d connected\n",i);

[saiyaman5]

服务器进入Sleep状态之后,
int nRet = connect( ConnectSocket, (SOCKADDR*) &clientService, sizeof(clientService) );
都是返回0

[thirddata]

可能是你的电脑上有另外一个程序使用6000端口，并且接受了你的连接，你应该检查一下你的服务端的程序bind和listen的返回值，是否是成功的。