java - TCP 连接，serversocket.accept 运行时不会创建套接字

Question

这是我的代码，当我在 Eclipse 中以 Debug模式运行它时，它显示它不会继续，它会停留在我放置箭头的代码中。

    private ServerSocket serverSocket = null;
    private Socket socket= null;
    private ObjectInputStream inputStream= null;    
public void ConnectTCP(){
        try{
            serverSocket = new ServerSocket(5000);
       ---->socket = serverSocket.accept();
            inputStream = new ObjectInputStream(socket.getInputStream());
            System.out.print("Server is Running");
        }catch(IOException e){
            e.printStackTrace();
        }
    }

Answer 1

您的套接字已在这一行创建。因为服务器绑定(bind)到一个端口，所以此时 ServerSocket 构造函数被调用。至于接受方法，due to JavaDoc它

Listens for a connection to be made to this socket and accepts it. The method blocks until a connection is made. A new Socket s is created and, if there is a security manager, the security manager's checkAccept method is called with s.getInetAddress().getHostAddress() and s.getPort() as its arguments to ensure the operation is allowed. This could result in a SecurityException.

所以，accept方法只是在等待客户端连接，这就是执行在此时停止的原因。也许，阅读 java official tutorial 可能会有所帮助。用于编写服务器端。