Hello,

I'm currently programming the movement engine on my multiplayer game. The game is played through a server which validates everything and sends updates to other clients.
Now I'm at the point where it has to validate player movement and send updates to all clients seeing that player.

Now my question is how I should process movement packets from client. What I thought of: Having one extra thread which does movement validation. This thread has a queue which stores incoming movement packets based on the order in which they arrive. Then, in every iteration, it takes the next movement packet from the queue and validates it.

Also of note: the game has tile based movement. Meaning one movement advances the player one tile further.

Is this an OK approach or is there a better one?

I guess I'll use the mac address solution. Any objections to this?

Also, how could I make a KryoNet client send his mac address during connect (not after the connection is established, during establishment)?

That is a given. Not allowing to log into the same account more than once.

But I also want to prevent the same person playing with more than one account on the same server (at least on the same machine).

Ok, but I would want to remove disconnected connections from the list. As far as I know you can't grab their IP/host anymore when they're disconnected, can you?

In that case I would need to iterate over the whole list and check for the disconnected connection id?

Hello,

I am using KryoNet to implement multiplayer functionality in my game. Now I don't want the same person to play more than once on the same server. Or to describe it a bit different: I don't want the same person have more than one connection established to the server.

Now I know you can't always suppress this. When someone is on different hosts/machines, he can do that. But at least I want to suppress more than one connection per machine.

How can I do that using KryoNet? Does KryoNet have such a functionality or do I have to code it myself?

I would be glad if you could share your wisdom with me.