I have several programs listening to the same multicast stream, I'm wondering will this doubling the traffic compared with only one program listening or the traffic/bandwidth usage are the same? thanks!
The short answer is no, the amount of traffic is the same. I'll caveat that with "in most cases". Multicast packets are written to the wire using a MAC address constructed from the multicast group address. Joining a multicast group is essentially telling the NIC to listen to the appropriate MAC address. This makes each listener receive the same ethernet frame. The caveat has to do with how multicast routing may or may not work. If you have a multicast aware router then multicast traffic may traverse the router onto other networks if someone has joined the group on another subnet.
I recommend reading "TCP/IP Illustrated, Volume 1" if you plan on doing a lot of network programming. This is the best way to really understand how all of the protocols fit together.
Are the clients on the same network?
For wireless 802.11 multicast, it depends on the implementation of Multicast at the wireless access point.
Some wireless access points do multicast to unicast conversion at the datalink layer and thus send a data separately to EACH client that has joined the multicast group.
If the AP is not doing unicast conversion, generally, your network utilization does not increase.
Related
I want to build an overlay network through VXLAN multicast to achieve communication between virtual machines, but I found that multicast packets can only be transmitted on the same subnet. In order to allow virtual machines on hosts that are not on the same subnet to communicate, I am thinking whether "capture and forward packets" would work. That is, grab a UDP packet with a destination address of 239.1.1.1 and a port of 4789 on hostA on network1, and send it to hostB on network2, and let hostB send the multicast packet. Then I found that the hosts on network2 all can catch this packet with wireshark, but no host corresponds to it. I wonder if Linux has a mechanism to discard fake multicast packets? If this is the case, how should this mechanism be prevented?
Unhandled, multicast is essentially broadcast. For IPv4 multicast that broadcast effect can be mitigated with IGMP. On switched networks with semi-intelligent switches there may be IGMP snooping functionality to further aid in this. Provided this exists, an end-device must subscribe to a multicast group by sending an IGMP join for the given group to "unfilter" that traffic towards itself. Routing multicast between subnets can be done with PIM or DVRMP implementations, or even static multicast routing daemons.
Only exception to this filtering is the 224.0.0.x range, which is reserved for link-local communication, usually IETF protocols. Traffic to these groups must never be filtered in any way.
Hence, to prevent filtering, either the end devices join the group (recommended!), or you send traffic to a group in the reserved range, e.g. on 224.0.0.1 the all-hosts group. (It's ugly and you may trigger ugly bugs on devices in the LAN, but it works.)
If I understood correctly, in multicast scenario, packets send to a multicast group are received only by the clients who joined the group(essentially the router forwards the packets ONLY to intended recipients). Then how is wireshark able to capture the multicast traffic(even in non promiscuous mode)?
Please help me understand if I am missing something. In the below sample capture, 192.168.1.1 is the server sending the multicast packets to group 239.0.0.222 and the current machine/wireshark did not join the group but still be able to capture the packets.
Topology:
I have 3 VMs, 1 Server, 1 Client and 1 WireShark VM. All of these three VMs are behind a Hyper-V Private switch(in the same subnet).
It's true that routers will only forward multicast traffic if there are clients on the other side that are expecting that traffic. But in your case all 3 VMs are in the same subnet so there's no router involved, only a switch.
The switch that the 3 VMs are connected to probably doesn't perform any special handing of multicast messages. This means that any multicast message it receives is being sent out on all ports, which includes the VM running Wireshark. And because the packets are physically arriving on that port, Wireshark sees them and displays them.
I am using this code to experiment with multicast.
I have wireshark running on another PC in the LAN and I am trying to see the packets sent over the network.
If packets are sent to the IP address 224.0.0.1 they are correctly received, if I send them to the address 239.255.255.250 they are not.
I have this entry in my routing table
224.0.0.0/4 dev eth0.100 src 192.0.0.1
I know 192.0.0.1 looks weird, but it is the correct interface address
Do you know why packets are not being sent?
The PC that you are attempting to capture the multicast traffic most likely has multicast filtering enabled and has not enabled reception of the 239.255.255.250 group.
Any host device that communicates multicast joins the 224.0.0.1 (All Hosts group) so packets sent to that group pass the filter, but not packets sent to 239.255.255.250.
Multicast filtering is usually implemented in the network interface and received frames that don't pass the filter will be dropped in hardware instead of being passed up to the Kernel driver. If the frames are not passed to the Kernel, then Wireshark can not capture them.
You can verify the active multicast groups per interface with netstat. I'm assuming you're on a Linux system since you tagged the question with "Linux"
netstat -g
Multicast groups are joined when a socket enables the IP_ADD_MEMBERSHIP socket option. In order to join the group on your capturing device, you'll need a program that creates a socket and joins the multicast group using the IP_ADD_MEMBERSHIP socket option
To add to Joel Cunningham's answer in the case of switched LAN:
There's nothing listening for multicast on 239.255.255.250, thus no IGMP joins are sent out. thus the switch does not propagate these multicast frames to the port where you have the wireshark(1)-running PC.
Again, 226.0.0.1 is well known and is always joined to by all hosts.
I am trying to list all opened multicast sockets on a linux system? netstat -g lists the groups joined though. Is there any other utility that I can use for this sake?
Thanks a lot for the help.
In addition to netstat -g you can use this to see all sockets which are bound to a multicast address:
netstat -anu|sort -nk4
This is a list of all UDP sockets (whether multicast or not). Look for all addresses in the range 224.0.0.0 to 239.255.255.255. These are sockets bound to multicast addresses, regardless whether they joined the multicast group or not. These will only receive traffic for this multicast group.
But:
In practice UDP sockets which are used to receive multicast traffic are usually bound to address 0.0.0.0. These can receive UDP packets for all unicast and multicast addresses, and the server usually does some additional filtering based on the source-IP-address.
So in that list above you may want to look also at UDP sockets bound to 0.0.0.0, for example 0.0.0.0:5353 which is most likely the mDNS (avahi, zeroconf) server.
I don't think there is a tool that can give you that information.
The reason is that a multicast socket is not bound to an address, it only participates in a multicast group (IP_ADD_MEMBERSHIP). A socket can join multiple different groups on the same interface, or same group on different interfaces, so it would make little sense to maintain these cross refences. The only information exposed by the kernel is in /proc/net, and in this case /proc/net/igmp (<- netstat -g).
I would like to learn that once and for all. What is the procedure to connect a multicast socket? I know you have to bind to a local interface (do you need IP and port for that?) then I know you have to join a group (do you need IP:PORT for the address you want to join and the network interface again!!!??) and then finally you can leave the group.
Can someone with experience clarify what is the whole of those many addresses? I will list below:
BindAddress (IP:PORT)
NetworkAddress (IP:PORT)
MulticastAddress (IP:PORT)
Where and what is the multicast group here?
A multicast group is a special IP address. You join it via setsockopt() using the socket option IP_ADDMEMBERSHIP, or e.g. in Java via MulticastSocket.joinGroup(). No port number here. If you want to join via a specific local address, use the overload that specifies a local address, or call setNetworkInterface() first.
Binding to a local address is a separate operation, which primarily determines which local addresses the socket can send and receive data on: one, or all of them: either one local address, which determines which of your available subnets you are listening to and can send via, or a port, or both. Normally it is best to use INADDR_ANY as the bind-address, unless your application magically knows about the network topology.
This is muddied by the fact that you can bind to a multicast address in Linux, but this appears to be a misunderstanding that will now always be with us.
You send to a multicast group by sending to the multicast address.
Yes, you must define both the address and the port for sending/receiving multicast messages. These are UDP packets, so they require both address and port for the networking stack to be able to properly deliver the messages to participating processes. So to listen for a particular set of multicast messages, your application needs to bind to a particular multicast ip address and port combination (and obviously for a set or all interfaces on the machine). The group is defined by the address/port combination.
Good quick explanation
Some sample source code in C and other languages