多代理问题
假设您有多个代理（每个代理都是它自己的进程），它们可以相互通信（代理只能成对通信）；如何确保一旦两个代理开始通信，没有其他代理可以中断他们？
这是代码的重要部分：
class Agent {
private void send(int to, byte[] message) {...};
private void receive(int from, byte[] message) {...};
}
send 方法将消息发送到指定的代理，而 receive 方法处理收到的来自其他代理的消息。因此，假设 id=1 和 id=2 的代理交换一些消息：如何确保两个代理在交换期间都不会处理（在其接收方法中）来自任何其他代理的消息？我尝试根据代理的 id 过滤消息（通过将它们存储在 int 变量中），但它似乎无法正常工作？
say you have multiple agents(each of them is it's own process) which can communicate to each other(agents communicate only in pairs); how do you ensure that once two agents start communicating, no other agent can interrupt them?
Here's the important part of the code:
class Agent {
private void send(int to, byte[] message) {...};
private void receive(int from, byte[] message) {...};
}
The send method sends the message to the specified agent and receive method processes the received message, that comes from other agents. So let's say agents with id=1 and id=2 exchange a few messages: how do I ensure that neither of the agents processes(in it's receive method) messages from any other agent during their exchange? I tried filtering messages based on agent's id(by storing them in int variable) but it doesn't seems to work properly?
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通过进程并使用 TCP 进行 IPC，任何时候让每个代理最多打开一个套接字。那么第二个代理将永远无法与已经通话的代理通话。
With processes and using TCP for the IPC, at any time let every agent have at most one socket open. Then a second agent will never be able to talk with an already talking agent.
您可以给接收者一个令牌。当代理想要发送给接收者时，它应该首先获取接收者的令牌。整个通信结束后，发送方将释放令牌，然后其他人就可以向接收方发送令牌。
You can give receivers a token. When an agent want to send to a receiver, it should get the receiver's token at first. After finishing the whole communication, the sender will release the token, and then other's can send to the receiver.