问题描述
我正在编写一个仅使用一个套接字的简单套接字客户端.所以我想我可以为所有接收操作重用一个SocketAsyncEventArgs.只要连接插座,它就可以正常工作.当服务器断开连接时,客户端将在控制台中进入一个无限的垃圾邮件发送循环,垃圾邮件收集成功".
I'm writing a simple socket client which uses only one socket. So I figured I could reuse a single SocketAsyncEventArgs for all receive operations. It's working fine as long as the socket is connected. When the server disconnects, the client will enter a infinite receive loop spamming "Receive Success" in the console.
e.SocketError!= SocketError.当套接字断开连接时成功为真吗?
Shouldn't e.SocketError != SocketError.Success be true when the socket is disconnected?
这是代码的一部分:
private void Completed(object sender, SocketAsyncEventArgs e)
{
if (e.SocketError != SocketError.Success)
status = 0;
System.Console.WriteLine(e.LastOperation + " " + e.SocketError);
if (status == 1)
{
switch (e.LastOperation)
{
case SocketAsyncOperation.Connect:
ProcessConnect(e);
break;
case SocketAsyncOperation.Receive:
ProcessReceive(e);
break;
case SocketAsyncOperation.Send:
ProcessSend(e);
break;
default:
status = 0;
break;
}
}
if (status != 1)
CloseSocket();
}
private void ProcessReceive(SocketAsyncEventArgs e)
{
if (!socket.ReceiveAsync(e))
Completed(null, e);
}
推荐答案
如果为套接字设置了字节流,则可能会收到零字节的回调,这表明套接字已正常关闭,并且不再读取任何字节.只需在您的代码中进行检查即可,例如
If your socket is set up for a byte stream then you may get a callback with zero bytes, this indicates that a graceful closure of the socket has occurred and that no more bytes will ever be read. Simply check for this in your code, e.g.
private void ProcessReceive( SocketAsyncEventArgs e )
{
AsyncUserToken token = e.UserToken as AsyncUserToken;
if( e.ByteTransferred > 0 && e.SocketError == SocketError.Success )
{
// .. process normally ..
}
else
{
CloseClientSocket( e );
}
}
private void CloseClientSocket( SocketAsyncEventArgs e )
{
AsyncUserToken token = e.UserToken as AsyncUserToken;
try
{
token.Socket.Shutdown( SocketShutdown.Send );
}
// throws error if it's already closed
catch( Exception ) {}
token.Socket.Close();
}
当然,您还需要阻止客户端尝试使用套接字,但是我敢肯定您可以解决这一问题.
Of course you'll also need to stop your client from trying to use the socket, but I'm sure you can work that part out.
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