For a client socket, I can use bind() to bind it to a specific source Ip address to select a specific interface. Or I can use connect() directly then it will pick the source ip based on routing table.
I wonder can I bind a client socket to an ip not belongs to any interfaces ? E.g.: I have two interfaces:
eth0 : ip0
eth1 : ip1
(1) If I bind the client socket to ip2. Is this feasible ?
(2) If (1) is feasible, assuming client socket sent packets thru eth0. Then I configure the iptables in this client host, to forward all incoming packets to ip0 (eth0). In this case, if there are packets sent back from server side with destination ip address is ip2 (assuming this packet will reach my client host). Will my client socket receive the packet ?
Thanks in advance.
I don't really understand your question, but here goes:
For client sockets, you typically want the the OS and its routing table to pick the best interface for you using any available port. In which case, you bind to INADDR_ANY (0) and port 0. Or don't explicitly call bind at at all. Just call connect() and it will do the right thing.
If you need the client connection to occur through a specific interface, then bind the socket to a specific IP address. And then the OS will attempt to use that interface for the subsequent connect call and all traffic after that.
Attempting to bind the socket to an IP that doesn't belong to a local interface is surely going to result in an error.
Not sure what you mean about the iptables stuff. Sounds dicey.
Please have a look at:
https://www.rsyslog.com/doc/v8-stable/configuration/modules/omfwd.html#ipfreebind
MAN:
https://man7.org/linux/man-pages/man7/ip.7.html
IP_FREEBIND (since Linux 2.4)
If enabled, this boolean option allows binding to an IP
address that is nonlocal or does not (yet) exist. This
permits listening on a socket, without requiring the
underlying network interface or the specified dynamic IP
address to be up at the time that the application is
trying to bind to it. This option is the per-socket
equivalent of the ip_nonlocal_bind /proc interface
described below.
I need to accept only connections from particular client ip address at server side. Should not use acl. With help of socket strict bind at server side can i do?
Example:
client ip address: 1.1.1.1
server ip address: 1.1.1.2
At server side:
1. Open a socket
2. Bind socket with 1.1.1.1(client ip address) with port no.
Will i be allow to do the second step at server side? Any special options are there to do?
Please let me know.
Thanks,
Boobesh
You can only bind the server port to an ip address to specify the interface to use.
For example your server has two network interfaces, one connected to the internet and one to a configuration network. The webserver should maybe only listen on the internet interface and a management tool only listen on the configuration network.
For your purpose you can accept the connection, compare the ip address and if it is not in the list of allowed clients close the connection immediately (or after sending an error message).
The other solution would be to use a firewall that is configured to allow only connections from the specified clients to the server port.
I agree with the friend above, u can only manage the ip and port in you server, but not client. u should compare the coming socket with the one u store in your server.
Can two applications on the same machine bind to the same port and IP address? Taking it a step further, can one app listen to requests coming from a certain IP and the other to another remote IP?
I know I can have one application that starts off two threads (or forks) to have similar behavior, but can two applications that have nothing in common do the same?
The answer differs depending on what OS is being considered. In general though:
For TCP, no. You can only have one application listening on the same port at one time. Now if you had 2 network cards, you could have one application listen on the first IP and the second one on the second IP using the same port number.
For UDP (Multicasts), multiple applications can subscribe to the same port.
Edit: Since Linux Kernel 3.9 and later, support for multiple applications listening to the same port was added using the SO_REUSEPORT option. More information is available at this lwn.net article.
Yes (for TCP) you can have two programs listen on the same socket, if the programs are designed to do so. When the socket is created by the first program, make sure the SO_REUSEADDR option is set on the socket before you bind(). However, this may not be what you want. What this does is an incoming TCP connection will be directed to one of the programs, not both, so it does not duplicate the connection, it just allows two programs to service the incoming request. For example, web servers will have multiple processes all listening on port 80, and the O/S sends a new connection to the process that is ready to accept new connections.
SO_REUSEADDR
Allows other sockets to bind() to this port, unless there is an active listening socket bound to the port already. This enables you to get around those "Address already in use" error messages when you try to restart your server after a crash.
Yes.
Multiple listening TCP sockets, all bound to the same port, can co-exist, provided they are all bound to different local IP addresses. Clients can connect to whichever one they need to. This excludes 0.0.0.0 (INADDR_ANY).
Multiple accepted sockets can co-exist, all accepted from the same listening socket, all showing the same local port number as the listening socket.
Multiple UDP sockets all bound to the same port can all co-exist provided either the same condition as at (1) or they have all had the SO_REUSEADDR option set before binding.
TCP ports and UDP ports occupy different namespaces, so the use of a port for TCP does not preclude its use for UDP, and vice versa.
Reference: Stevens & Wright, TCP/IP Illustrated, Volume II.
In principle, no.
It's not written in stone; but it's the way all APIs are written: the app opens a port, gets a handle to it, and the OS notifies it (via that handle) when a client connection (or a packet in UDP case) arrives.
If the OS allowed two apps to open the same port, how would it know which one to notify?
But... there are ways around it:
As Jed noted, you could write a 'master' process, which would be the only one that really listens on the port and notifies others, using any logic it wants to separate client requests.
On Linux and BSD (at least) you can set up 'remapping' rules that redirect packets from the 'visible' port to different ones (where the apps are listening), according to any network related criteria (maybe network of origin, or some simple forms of load balancing).
Yes Definitely. As far as i remember From kernel version 3.9 (Not sure on the version) onwards support for the SO_REUSEPORT was introduced. SO_RESUEPORT allows binding to the exact same port and address, As long as the first server sets this option before binding its socket.
It works for both TCP and UDP. Refer to the link for more details: SO_REUSEPORT
No. Only one application can bind to a port at a time, and behavior if the bind is forced is indeterminate.
With multicast sockets -- which sound like nowhere near what you want -- more than one application can bind to a port as long as SO_REUSEADDR is set in each socket's options.
You could accomplish this by writing a "master" process, which accepts and processes all connections, then hands them off to your two applications who need to listen on the same port. This is the approach that Web servers and such take, since many processes need to listen to 80.
Beyond this, we're getting into specifics -- you tagged both TCP and UDP, which is it? Also, what platform?
You can have one application listening on one port for one network interface. Therefore you could have:
httpd listening on remotely accessible interface, e.g. 192.168.1.1:80
another daemon listening on 127.0.0.1:80
Sample use case could be to use httpd as a load balancer or a proxy.
When you create a TCP connection, you ask to connect to a specific TCP address, which is a combination of an IP address (v4 or v6, depending on the protocol you're using) and a port.
When a server listens for connections, it can inform the kernel that it would like to listen to a specific IP address and port, i.e., one TCP address, or on the same port on each of the host's IP addresses (usually specified with IP address 0.0.0.0), which is effectively listening on a lot of different "TCP addresses" (e.g., 192.168.1.10:8000, 127.0.0.1:8000, etc.)
No, you can't have two applications listening on the same "TCP address," because when a message comes in, how would the kernel know to which application to give the message?
However, you in most operating systems you can set up several IP addresses on a single interface (e.g., if you have 192.168.1.10 on an interface, you could also set up 192.168.1.11, if nobody else on the network is using it), and in those cases you could have separate applications listening on port 8000 on each of those two IP addresses.
Just to share what #jnewton mentioned.
I started an nginx and an embedded tomcat process on my mac. I can see both process runninng at 8080.
LT<XXXX>-MAC:~ b0<XXX>$ sudo netstat -anp tcp | grep LISTEN
tcp46 0 0 *.8080 *.* LISTEN
tcp4 0 0 *.8080 *.* LISTEN
Another way is use a program listening in one port that analyses the kind of traffic (ssh, https, etc) it redirects internally to another port on which the "real" service is listening.
For example, for Linux, sslh: https://github.com/yrutschle/sslh
If at least one of the remote IPs is already known, static and dedicated to talk only to one of your apps, you may use iptables rule (table nat, chain PREROUTING) to redirect incomming traffic from this address to "shared" local port to any other port where the appropriate application actually listen.
Yes.
From this article:
https://lwn.net/Articles/542629/
The new socket option allows multiple sockets on the same host to bind to the same port
Yes and no. Only one application can actively listen on a port. But that application can bequeath its connection to another process. So you could have multiple processes working on the same port.
You can make two applications listen for the same port on the same network interface.
There can only be one listening socket for the specified network interface and port, but that socket can be shared between several applications.
If you have a listening socket in an application process and you fork that process, the socket will be inherited, so technically there will be now two processes listening the same port.
I have tried the following, with socat:
socat TCP-L:8080,fork,reuseaddr -
And even though I have not made a connection to the socket, I cannot listen twice on the same port, in spite of the reuseaddr option.
I get this message (which I expected before):
2016/02/23 09:56:49 socat[2667] E bind(5, {AF=2 0.0.0.0:8080}, 16): Address already in use
If by applications you mean multiple processes then yes but generally NO.
For example Apache server runs multiple processes on same port (generally 80).It's done by designating one of the process to actually bind to the port and then use that process to do handovers to various processes which are accepting connections.
Short answer:
Going by the answer given here. You can have two applications listening on the same IP address, and port number, so long one of the port is a UDP port, while other is a TCP port.
Explanation:
The concept of port is relevant on the transport layer of the TCP/IP stack, thus as long as you are using different transport layer protocols of the stack, you can have multiple processes listening on the same <ip-address>:<port> combination.
One doubt that people have is if two applications are running on the same <ip-address>:<port> combination, how will a client running on a remote machine distinguish between the two? If you look at the IP layer packet header (https://en.wikipedia.org/wiki/IPv4#Header), you will see that bits 72 to 79 are used for defining protocol, this is how the distinction can be made.
If however you want to have two applications on same TCP <ip-address>:<port> combination, then the answer is no (An interesting exercise will be launch two VMs, give them same IP address, but different MAC addresses, and see what happens - you will notice that some times VM1 will get packets, and other times VM2 will get packets - depending on ARP cache refresh).
I feel that by making two applications run on the same <op-address>:<port> you want to achieve some kind of load balancing. For this you can run the applications on different ports, and write IP table rules to bifurcate the traffic between them.
Also see #user6169806's answer.