socket API接受（）函数如何工作？(How does the socket API accept() function work?)

 套接字API是TCP / IP和UDP / IP通信的实际标准（即我们所知道的网络代码）。 但是，它的一个核心功能是accept()有点神奇。 
 
 借用半正式定义： 
 
 
 
 在服务器端使用accept（）。 它接受从远程客户端创建新的TCP连接的接收到的尝试，并创建与此连接的套接字地址对相关联的新套接字。 
 
 
 换句话说， accept返回一个新的套接字，服务器可以通过这个新的套接字与新连接的客户端通信。 在同一个端口上，旧套接字（在其上被称为接受者）保持打开状态，监听新连接。 
 
 accept工作怎么办？ 如何实施？ 这个话题有很多混乱。 许多人声称接受打开一个新的端口，并通过它与客户端进行通信。 但这显然不是真的，因为没有新的港口被打开。 您实际上可以通过与不同客户端的同一个端口进行通信，但是如何？ 当几个线程在同一个端口上调用recv时，数据如何知道去哪里？ 
 
 我猜这是客户端的地址与一个套接字描述符相关的东西，每当数据通过recv它被路由到正确的套接字，但我不确定。 
 
 对这种机制的内在作用进行彻底的解释是非常好的。 

The socket API is the de-facto standard for TCP/IP and UDP/IP communications (that is, networking code as we know it). However, one of its core functions, accept() is a bit magical.
 
To borrow a semi-formal definition:
 
 
 
accept() is used on the server side. It accepts a received incoming attempt to create a new TCP connection from the remote client, and creates a new socket associated with the socket address pair of this connection.
 
 
In other words, accept returns a new socket through which the server can communicate with the newly connected client. The old socket (on which accept was called) stays open, on the same port, listening for new connections.
 
How does accept work? How is it implemented? There's a lot of confusion on this topic. Many people claim accept opens a new port and you communicate with the client through it. But this obviously isn't true, as no new port is opened. You actually can communicate through the same port with different clients, but how? When several threads call recv on the same port, how does the data know where to go?
 
I guess it's something along the lines of the client's address being associated with a socket descriptor, and whenever data comes through recv it's routed to the correct socket, but I'm not sure.
 
It'd be great to get a thorough explanation of the inner-workings of this mechanism.

原文：https://stackoverflow.com/questions/489036