What is a TCP window update?

phpscriptcoder picture phpscriptcoder · Sep 23, 2009 · Viewed 52.3k times · Source

I'm making my own custom server software for a game in Java (the game and original server software were written with Java). There isn't any protocol documentation available, so I am having to read the packets with Wireshark.

While a client is connecting the server sends it the level file in Gzip format. At about 94 packets into sending the level, my server crashes the client with an ArrayIndexOutOfBoundsException. According to the capture file from the original server, it sends a TCP Window Update at about that point. What is a TCP Window Update, and how would I send one using a SocketChannel?

Answer

user73774 picture user73774 · Sep 23, 2009

TCP windows are used for flow control between the peers on a connection. With each ACK packet, a host will send a "window size" field. This field says how many bytes of data that host can receive before it's full. The sender is not supposed to send more than that amount of data.

The window might get full if the client isn't receiving data fast enough. In other words, the TCP buffers can fill up while the application is off doing something other than reading from its socket. When that happens, the client would send an ACK packet with the "window full" bit set. At that point, the server is supposed to stop sending data. Any packets sent to a machine with a full window will not be acknowledged. (This will cause a badly behaved sender to retransmit. A well-behaved sender will just buffer the outgoing data. If the buffer on the sending side fills up too, then the sending app will block when it tries to write more data to the socket!)

This is a TCP stall. It can happen for a lot of reasons, but ultimately it just means the sender is transmitting faster than the receiver is reading.

Once the app on the receiving end gets back around to reading from the socket, it will drain some of the buffered data, which frees up some space. The receiver will then send a "window update" packet to tell the sender how much data it can transmit. The sender starts transmitting its buffered data and traffic should flow normally.

Of course, you can get repeated stalls if the receiver is consistently slow.

I've worded this as if the sender and receiver are different, but in reality, both peers are exchanging window updates with every ACK packet, and either side can have its window fill up.

The overall message is that you don't need to send window update packets directly. It would actually be a bad idea to spoof one up.

Regarding the exception you're seeing... it's not likely to be either caused or prevented by the window update packet. However, if the client is not reading fast enough, you might be losing data. In your server, you should check the return value from your Socket.write() calls. It could be less than the number of bytes you're trying to write. This happens if the sender's transmit buffer gets full, which can happen during a TCP stall. You might be losing bytes.

For example, if you're trying to write 8192 bytes with each call to write, but one of the calls returns 5691, then you need to send the remaining 2501 bytes on the next call. Otherwise, the client won't see the remainder of that 8K block and your file will be shorter on the client side than on the server side.