意见不同,但我肯定会为每个客户端使用一个线程。然后,与单个处理线程的通信可以通过 LinkedBlockingQueue,或者只是一个同步的 LinkedList。
在每个客户端线程上是这样的:
public class Client implements Runnable, ResponseOutput {
private final BufferedReader br;
private final PrintWriter pw;
public Client(Socket s) {
br = new BufferedReader(new InputStreamReader(s.getInputStream()));
pw = new PrintWriter(s.getOutputStream());
}
// defined by the ResponseOutput interface
public void sendReply(String reply) {
pw.println(reply);
}
public void run() {
try {
while (true) {
String s = br.readLine();
if (s==null)
break;
Processor.queue(new Processor.InputItem(this, s));
}
} catch (IOException ioe) {
... error handling ...
}
}
}
然后进行处理:
public class Processor implements Runnable {
static public class InputItem {
final ResponseOutput client;
final String command;
public InputItem(ResponseOutput client, String command) {
this.client = client;
this.command = command;
}
}
static private Processor instance;
static public void queue(InputItem item) {
instance.commandQueue.add(item);
}
private BlockingQueue<InputItem> commandQueue;
public void run() {
try {
while (true) {
InputItem item = commandQueue.take();
String reply = doStuff(item.command);
item.client.sendReply(reply);
}
} catch (InterruptedException ie) {
... error handling ....
}
}
}
在InputItem
类中,您还可以包含对需要更新的任何游戏状态的引用。由于只有处理线程在更改它,因此您无需任何同步即可执行此操作。