我正在学习套接字,目前;我正在努力向所有连接的客户端广播消息.
服务器只是用以下代码启动:
public static int Main(String[] args)
{
Thread t1 = new Thread(Test);
t1.Start();
#region Start listening to socket
IPHostEntry ipHostInfo = Dns.GetHostEntry(Dns.GetHostName());
IPAddress ipAddress = ipHostInfo.AddressList[0];
IPEndPoint localEndPoint = new IPEndPoint(ipAddress, 11000);
Socket listener = new Socket(ipAddress.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
try
{
listener.Bind(localEndPoint);
listener.Listen(100);
Console.WriteLine("[{0}] Server started listening!", DateTime.Now);
while (true)
{
// Set the event to nonsignaled state.
allDone.Reset();
// Start an asynchronous socket to listen for connections.
listener.BeginAccept(new AsyncCallback(AcceptCallback), listener);
// Wait until a connection is made before continuing.
allDone.WaitOne();
}
}
catch (Exception e)
{
Console.WriteLine(e.ToString());
}
#endregion
Console.WriteLine("\nPress ENTER to continue...");
Console.Read();
return 0;
}
在那里,我启动套接字以侦听是否有人发送消息.
如您所见,我在那里建立了线程,并且希望该线程不时向所有连接的用户发送一些消息(目前所有时间仅几个字节用于测试).
这是线程中的方法
private static void Test()
{
while (true)
{
for (int i = 0; i < Players.Length; i++)
{
if (Players[i] == null)
continue;
Socket cs = new Socket(Players[i].IPEndPoint.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
cs.Bind(Players[i].IPEndPoint);
byte[] data = Encoding.ASCII.GetBytes("Random data");
cs.Send(data);
}
}
}
此方法仅是最后一次尝试,但未成功.问题是它使我发现以下错误:
socket is reference of null.
在这种情况下,但无论如何,我都无法将消息发送给每个客户端.
这是我跟踪客户端与服务器连接的方式.
case "0": // Client tries to connect to server
if (nClients >= MAX_PLAYERS)
{
Send(handler, "Too many players. Try again later.");
return;
}
for (int i = 0; i < MAX_PLAYERS; i++)
{
if(Players[i] == null)
{
Players[i] = new Client();
Players[i].Name = parts[1].ToString();
Players[i].IPEndPoint = handler.RemoteEndPoint as IPEndPoint;
Send(handler, String.Format("1|{0}", i));
Console.WriteLine("[{0}] Succesfully registered client ID: {1}, NAME: {2}!", DateTime.Now, i, parts[1].ToString());
i = MAX_PLAYERS;
nClients++;
}
}
break;
这只是专注于处理连接消息的代码的一部分,该消息位于我的方法内部:
private static void HandleMessages(string message, Socket handler)
我从这里调用它:
public static void AcceptCallback(IAsyncResult ar)
{
// Signal the main thread to continue.
allDone.Set();
// Get the socket that handles the client request.
Socket listener = (Socket)ar.AsyncState;
Socket handler = listener.EndAccept(ar);
// Create the state object.
StateObject state = new StateObject();
state.workSocket = handler;
handler.BeginReceive(state.buffer, 0, StateObject.BufferSize, 0, new AsyncCallback(ReadCallback), state);
}
public static void ReadCallback(IAsyncResult ar)
{
String content = String.Empty;
// Retrieve the state object and the handler socket
// from the asynchronous state object.
StateObject state = (StateObject)ar.AsyncState;
Socket handler = state.workSocket;
// Read data from the client socket.
int bytesRead = handler.EndReceive(ar);
if (bytesRead > 0)
{
// There might be more data, so store the data received so far.
state.sb.Append(Encoding.ASCII.GetString(
state.buffer, 0, bytesRead));
// Check for end-of-file tag. If it is not there, read
// more data.
content = state.sb.ToString();
if (content.IndexOf("<EOF>") > -1)
{
string message = content.Remove(content.Length - 5, 5);
HandleMessages(message, handler);
}
else
{
// Not all data received. Get more.
handler.BeginReceive(state.buffer, 0, StateObject.BufferSize, 0,
new AsyncCallback(ReadCallback), state);
}
}
}
解决方法:
虽然我无法为您编写所有代码,但我会教您一些可以使您步入正轨的模式.
我假设您正在通过TCP进行通信.套接字代表一个连接.如果要发送内容,则必须使用与接收相同的Socket实例.创建一个新的套接字并绑定将用于打开另一个监听端口(此处不适用).
可能希望每个连接都具有一个对象,以跟踪所有相关状态.例如:
class MyConnection {
Socket socket;
Player player;
}
在List< MyConnection>中跟踪该类的实例.然后,您可以遍历该列表并使用套接字发送内容.
您可以简化很多接受循环.这是一个很好的模式:
while (true) {
var connectionSocket = listeningSocket.Accept();
Task.Run(() => ProcessConnection(connectionSocket));
}
Microsoft的示例代码中的所有异步内容都没有用.只有一个接受线程.异步用于保存线程.保存一个线程没有帮助.而且,阻止事件完全抵消了异步的好处.
如果连接的客户端数量少(< 100),并且/或者如果您关心简单的代码,则可以简单地摆脱所有异步IO并使用同步IO(无Begin / End). 或者,您可以对套接字使用基于任务的包装器,也可以使用具有基于任务的方法的NetworkStream.然后,您可以使用await.这也摆脱了那些使编写简单逻辑变得非常困难的回调. 如果不足以帮助您,我们可以在评论中进行处理.