ServerSocket 是否接受任意端口上的返回套接字?

本文介绍了ServerSocket 是否接受任意端口上的返回套接字?的处理方法,对大家解决问题具有一定的参考价值

问题描述

关于 java 中的 serversockets,我已经看到很多与此类似的答案:假设您有一台服务器,在端口 5000 上有一个 serversocket.客户端 A 和客户端 B 将连接到我们的服务器.

I've seen many answers similar to this one in regards to serversockets in java: "Let's say you have a server with a serversocket on port 5000. Client A and Client B will be connecting to our server.

客户端 A 在端口 5000 上向服务器发送请求.客户端 A 端的端口由操作系统选择.通常,操作系统会选择下一个可用的端口.此搜索的起点是先前使用的端口号 + 1(例如,如果操作系统最近发生在我们的端口 45546 上,则操作系统将尝试 45547).

Client A sends out a request to the Server on port 5000. The port on Client A's side is chosen by the Operating System. Usually, the OS picks the next port that is available. The starting point for this search is the previously-used port number + 1 (so for instance if the OS happened to us port 45546 recently, the OS would then try 45547).

假设没有连接问题,服务器收到客户端 A 的连接请求,连接到端口 5000.服务器然后打开自己的下一个可用端口,并将其发送给客户端.在这里,客户端 A 连接到新端口,服务器现在可以再次使用端口 5000."

Assuming there are no connection problems, the Server receives Client A's request to connect on port 5000. The Server then opens up its own next available port, and sends that to the client. Here, Client A connects to the new port, and the server now has port 5000 available again."

我在关于 stackoverflow 的多个问题中看到了这样的答案,这些问题是关于在 accept() 返回的套接字中如何使用不同的端口而不是 ServerSocket 正在侦听的端口.我一直以为TCP是由四方信息识别的:

I've seen answers like this in multiple questions on stackoverflow about how a different port is used in the returned socket of the accept() than the port that the ServerSocket is listening on. I was always under the impression that TCP is identified by the quartet of information:

Client IP : Client Port and Server IP : Server Port ->协议也是(区分TCP和UDP)

Client IP : Client Port and Server IP : Server Port ->protocol too (to distinguish TCP from UDP)

那么为什么accept() 需要返回一个绑定到不同端口的套​​接字呢?每个标头中发送的四重信息是否足以区分来自不同机器的同一服务器端口的多个连接,不需要使用服务器机器上的不同端口进行通信?

So why would the accept() need to return a socket bound to a different port? Doesn't the quartet of information sent in every header distinguish multiple connections to the same server port from different machines enough where it would not need to use different ports on the server machine for communication?

推荐答案

您在 TCP 数据包头信息中是正确的.它包含:

You are correct in the TCP packet header's information. It contains:

Client IP | Client Port | Server IP | Server Port | Protocol

或者,更合适的(因为当您考虑双向传输时客户端/服务器会变得混乱):

Or, more appropriately (since client/server become confusing when you think about bi-directional transfer):

Source IP | Source Port | Destination IP | Destination Port | Protocol

到同一服务器端口的多个连接将来自客户端上的不同端口.一个例子可能是:

Multiple connections to the same server port will come from different ports on the client. An example may be:

0.0.0.0:45000 -> 1.1.1.1:80
0.0.0.0:45001 -> 1.1.1.1:80

客户端端口的差异足以消除两个套接字的歧义,因此具有两个独立的连接.服务器不需要在另一个端口上打开另一个套接字.它确实从 accept 方法接收了一个套接字,但它被分配到同一个端口,现在是一条返回新接受的客户端的路由.

The difference in client ports is enough to disambiguate the two sockets, and thus have two separate connections. There is no need for the server to open another socket on another port. It does receive a socket from the accept method, but it's assigned to the same port and is now a route back to the newly accepted client.

另一方面,FTP 确实有一个模型,其中服务器将打开一个新的非特权端口(> 1023)并将其发送回客户端以供客户端连接(这称为被动 FTP").这是为了解决客户端位于防火墙后面并且无法接受来自服务器的传入数据连接的问题.但是,在典型的 HTTP 服务器(或任何其他标准套接字实现)中,情况并非如此.它的功能位于 FTP 之上.

FTP, on the other hand, does have a model where the server will open a new unprivileged port (> 1023) and send that back to the client for the client to connect to (this is referred to as "Passive FTP"). This is to resolve issues where the client is behind a firewall and can't accept incoming data connections from the server. However, this is not the case in a typical HTTP server (or any other standard socket implementation). It's functionality that is layered on top of FTP.

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