如何从外部连接到运行在127.0.0.1（而不是0.0.0.0）上的服务？

Question

I'm trying to connect to a service, and to debug it, I ran

netstat -nap | grep LISTEN

The results should rows of two types :

tcp 0 0 127.0.0.1:8020 0.0.0.0:*     LISTEN      
tcp 0 0 0.0.0.0:57140  0.0.0.0:*     LISTEN      
tcp 0 0 0.0.0.0:11000  0.0.0.0:*     LISTEN      
tcp 0 0 0.0.0.0:8088   0.0.0.0:*     LISTEN 
unix 2 [ ACC ]  STREAM LISTENING     4512   -                   
unix 2 [ ACC ]  STREAM LISTENING     9760   -                   

I have 3 questions :

1) I want to connect to the process running on 127.0.0.1 --- how can I do this externally ? I have read elsewhere that 127.0.0.1 processes are only allowed to communicate with other localhost processes.

2) What is the difference between the "tcp 0" netstat records and the "unix 2" ones ? Im somewhat naive about networking, so feel free to overexplain this one :)

Answer 1

In short, your process is bound to a loopback interface which cannot receive packets from an external network. You'll need to reconfigure the process bound to port 8020 to bind to an external interface to be able to connect to it from another host.

The long answer is that the two addresses you site (127.0.0.1 and 0.0.0.0) are both special in certain ways, and it is useful to understand what you're seeing.

Addresses in the 127.0.0.0/8 Internet Protocol address block (of which 127.0.0.1 is one) are reserved for use internally on a host. See rfc5735 for details, but there's nothing special about these addresses except that all IP hosts use the same rules and aren't setup to route these addresses outside a host or router.

On your computer, you'll usually see a special "loopback" network interface that has 127.0.0.1 assigned.

lo        Link encap:Local Loopback  
          inet addr:127.0.0.1  Mask:255.0.0.0

This interface is special and never connected to an external network. It is used when a program wants to connect to a service on the local machine as 127.0.0.1 will almost always be configured as an active network interface. Packets will only arrive on this interface if they are sent from a local process.

The other address you site, 0.0.0.0 is special and usually represents all IP addresses mapped to any network interface on your computer. When a program wants to listen for connections arriving on any network interface or IP address, it will bind a TCP/UDP port to 0.0.0.0 to listen for connections.

In your case, however, you're reporting netstat output listing 0.0.0.0 on lines describing TCP sockets in a LISTEN state. In this case, netstat is listing sockets listening for connections and using 0.0.0.0:* as a place holder for the foreign address field of it's output. In this case, 0.0.0.0:* signifies that the socket is waiting for a connection from any host.

Regarding your question on "tcp 0" vs. "unix 2", these are the first two columns of your netstat output. A look at the column headers from your netstat command is useful:

# netstat -nap | head -2
Active Internet connections (servers and established)
Proto Recv-Q Send-Q Local Address               Foreign Address             State       PID/Program name 

What you're reporting as "tcp 0" simply means a socket using the TCP protocol has zero bytes in the received queue waiting for the program connected to this socket to consume. Similarly, "unix 2" is what's called a unix socket with two bytes waiting in its receive queue for the connected process to consume.

TCP sockets are part of the TCP/IP stack that can be used locally or across IP networks for processes to communicate. UNIX sockets, on the other hand, are simpler and only used for what's called IPC or inter-process communication which only happens between two processes both running on the local system, and there's no networking involved (no addresses and ports anyway). UNIX sockets are considered to be more efficient than TCP sockets, but they are obviously more limited in function. On UNIX-like systems UNIX sockets are implemented as a file on the file system of a special "socket" type that both processes using the socket read and write to as a communication channel.

Answer 2

1) Without binding it to 0.0.0.0, you can still access the service through a tunnel. This is similar to using a proxy as David Schwartz mentioned. There's a few assumptions I'm making for this example:

1. The server is running a service bound to 127.0.0.1:8020, we'll call it 'myservice'.
2. The server is running OpenSSH server 'sshd' on the default port of TCP 22, and the user can log in with the username 'myusername'.
3. The client is running a system with OpenSSH client installed.
4. The server is accessible via the IP address of 10.20.30.40.

On the client, SSH to the server with the following command:

ssh -L 12345:localhost:8020 myusername@10.20.30.40

Once you log in, minimize the SSH window. In another window on the client, run netstat to find listening ports. You should see 127.0.0.1:12345, just like on the server.

On the client, connect to the service on 127.0.0.1:12345. You should now be connected to the 'myservice' instance on the server, even though you made the connection to the client's local loopback interface.

The trick here is that SSH is tunneling a listening socket on the client to the listening socket on the server. I've made the port numbers different for clarity.

Answer 3

1) You would either need to modify the server to bind to a publicly accessible address (or 0.0.0.0) or run a local proxy to handle the connection.

2) TCP connections use the TCP protocol, the one used for connection-oriented traffic on the Internet. UNIX connections use a strictly local protocol that is much simpler than TCP (because it doesn't have to deal with dropped packets, lost routes, corrupted data, out of order packets, and so on).

Answer 4

1) You cannot (if you mean from another machine - 127.0.0.1 is localhost and by definition you can only connect to it from the local machine

2) The first column shows the domain of the sockets - tcp are tcp sockets and unix are unix domain sockets.

And as for the answer to you question 3 ;-)

3) 42