I have a root node(server) connected to many other nodes(clients) through TCP sockets. I want to send some data from server to client, but the data is different for each node and depends on the ip address of that node.
Thus I should have ip address of each node connected to server. How can I have that information?
When you call accept(2) you can choose to retrieve the address of the client.
int accept(int socket, struct sockaddr *restrict address,
socklen_t *restrict address_len);
You need to store those addresses and then send(2) to each what you need to send.
So the workflow should be something like this:
Keep a list of connected clients. Initially the list is empty, of course
When you accept a connection, push its details into that list (the address and the socket returned by accept(2)).
When you need to send something to every client, simply walk the list and send it (using the stored socket)
The one tricky part is that socklen_t *restrict address_len is a value-result argument, so you need to be careful with that.
This is a more nuanced question than it first appears.
If the clients are sitting behind a NAT, you may get the same IP from more than one client. This is perfectly natural and expected behavior. If you need to distinguish between multiple clients behind the same NAT, you'll need some other form of unique client id (say, IP address and port).
As long as you have access to the list of file descriptors for the connected TCP sockets, it is easy to retrieve the addresses of the remote hosts. The key is the getpeername() system call, which allows you to find out the address of the remote end of a socket. Sample C code:
// This is ugly, but simpler than the alternative
union {
struct sockaddr sa;
struct sockaddr_in sa4;
struct sockaddr_storage sas;
} address;
socklen_t size = sizeof(address);
// Assume the file descriptor is in the var 'fd':
if (getpeername(fd, &address.sa, &size) < 0) {
// Deal with error here...
}
if (address.sa.family == AF_INET) {
// IP address now in address.sa4.sin_addr, port in address.sa4.sin_port
} else {
// Some other kind of socket...
}
Related
Can we detect different clients(devices) from same IP on TCPConnection ?
Example :
I have a TCP Server called s1 and I have 2 PCs called p1,p2 and my PCs IP is same (e.g. 1.2.3.4)
when I connect to s1 (my TCP Server) with p1 and p2 (Pc1 and Pc2) can s1(my TCP server) detect these clients with same IP, isn't same device ?
From my understanding you are basically asking to detect different devices behind a NAT, i.e. devices sharing the same external IP address. There is no fully reliable way to do this but one can employ heuristics. Typically these are based on the ID field in the IP header and/or the TCP timestamp option, see for example A Technique for Counting NATted Hosts or Time has something to tell us about Network
Address Translation. One might also try to use passive OS fingerprinting in order to detect if different OS are used (and thus different real or virtual devices) - see Passive Fingerprinting.
None of these heuristics are fully reliable though and they also will not work if the devices are behind a proxy, since in this case the TCP/IP connections visible to the server originate from a single device - the proxy.
Yes you can. The server can ask the operating system for the connection information of client associated with the socket. In 'C' this would look like:
//Accept and incoming connection
puts("Waiting for incoming connections...");
c = sizeof(struct sockaddr_in);
new_socket = accept(socket_desc, (struct sockaddr *)&client, (socklen_t*)&c);
if (new_socket<0)
{
perror("accept failed");
return 1;
}
The client sockaddr structure will be filled with the information about the connecting client. The server can look into this to extract the IP Address as a string doing something like:
char *ip = inet_ntoa((struct sockaddr_in *)client.sin_addr);
You can now see if ip matches p1 or p2.
I have a simulator application which Unix Domain datagram sockets, which sends data to socket path for.ex /var/lib/XYZ.
sendto is returning -2 which is due to other end no peer is there(no other unix domian socket application is running)
I would like to write a datagram client/peer application using Unix Domain Sockets for receiving data from the server/simulator(which is sending data to /var/lib/XYZ).
My code is as follows:
#define BUF_SIZE 1024
#define SV_SOCK_PATH "/var/lib/XYZ"
#define SV_SOCK_PATH2 "/var/lib/ABC"
creation of Unix Domain sockets as below:
struct sockaddr_un svaddr, claddr;
....
sfd = socket(AF_UNIX, SOCK_DGRAM, 0);
if (sfd == -1)
printf("socket creation failed");
memset(&claddr, 0, sizeof(struct sockaddr_un));
claddr.sun_family = AF_UNIX;
strncpy(claddr.sun_path, SV_SOCK_PATH2, sizeof(claddr.sun_path) - 1);
if (bind(sfd, (struct sockaddr *) &claddr, sizeof(struct sockaddr_un)) == -1)
printf("bind failed");
/* Construct address of server */
memset(&svaddr, 0, sizeof(struct sockaddr_un));
svaddr.sun_family = AF_UNIX;
strncpy(svaddr.sun_path, SV_SOCK_PATH, sizeof(svaddr.sun_path) - 1);
while(1)
{
int len=sizeof(struct sockaddr_un);
numBytes = recvfrom(sfd, resp, BUF_SIZE, 0, (struct sockaddr*)&svaddr,&len);
if (numBytes == -1)
printf("recvfrom error");
else{
printf("no of bytes received from server: %d",(int)numBytes);
printf("Response %d: %s\n", (int) numBytes, resp);
}
}
remove(claddr.sun_path);
//exit(EXIT_SUCCESS);
}
but the program is not receiving anything...is there anything missed out??
When it comes to datagrams, there is no real client or server. Either side attempting to send is responsible for addressing datagrams to the other. So, in your code, the setup is all wrong. You're apparently attempting to direct the "server" (but really not a server, just the other peer) to send to you via svaddr but that isn't how it works.
For a datagram AF_UNIX socket, the sender either needs to explicitly specify the receiver's address in a sendto call, or it needs to first connect its socket to the receiver's address. (In the latter case, it can then use send instead of sendto since the peer address has been specified via connect.)
You can't specify the sending peer's address in the recvfrom call. The socket address argument in the recvfrom is intended to return to you the address from which the datagram was sent. Whatever is in that variable will be overwritten on successful return from recvfrom.
One way datagram peer programs are often structured: the "server" creates a well-known path and binds to it, then a "client" creates its own endpoint and binds to it (constructing a unique socket address for itself), then the client can sendto the server's well-known socket. The server, by using recvfrom to obtain the client's address along with the datagram, can then use sendto along with the address to return a message to the client (without needing to connect its socket). This provides a sort of client-server paradigm on top of the fundamentally equal-peer orientation of the datagram socket.
Finally, I should mention that it's usually a good idea to use fully specified pathnames to ensure both peers are using the same address even if started from different directories. (Normally, with AF_UNIX, the address is a path name in the file system used to "rendezvous" between the two peers -- so without a full path "some_socket" is "./some_socket" in the current working directory. Some systems, such as linux, also support an abstract "hidden" namespace that doesn't require a full path, but you must use an initial null byte in the name to specify that.)
I am trying to send UDP packet to my server 10.20.1.2 with port number 20000. I have implemented UDP client on PC and when i send data using sendto API , at the same time i monitor data on wireshark wireshark shows it as an ARP packet.
18967 5440.858646 PcsCompu_ef:b4:89 Broadcast ARP 42 Who has 10.20.1.2? Tell 192.168.1.70
192.168.1.70 is my machine ip where UDP client is running.
I am not sure how UDP packet is getting converted into ARP packet ?
I understand ARP is for finding MAC address of target node but here i already know MAC address of target device , How can i add it in my udp client so it directly starts UDP communication . My target device is one embedded camera , i am not expecting it to reply on ARP request so i want to prevent sending ARP request.
Below is my UDP client code :
Any inputs are highly appreciated. Thanks in advance.
/*
Simple udp client
*/
#include<stdio.h> //printf
#include<string.h> //memset
#include<stdlib.h> //exit(0);
#include<arpa/inet.h>
#include<sys/socket.h>
#define SERVER "10.20.1.2"
#define PORT 20000 //The port on which to send data
char message[3]={0x00, 0x00 , 0x24};
int main(void)
{
struct sockaddr_in si_other;
int s, i, slen=sizeof(si_other);
int ret;
if ( (s=socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP)) == -1)
{
printf("socket failed");
}
memset((char *) &si_other, 0, sizeof(si_other));
si_other.sin_family = AF_INET;
si_other.sin_port = htons(PORT);
if (inet_aton(SERVER , &si_other.sin_addr) == 0)
{
fprintf(stderr, "inet_aton() failed\n");
exit(1);
}
ret = sendto(s, message, sizeof(message) , 0 , (struct sockaddr *) &si_other, slen);
close(s);
return 0;
}
Some clarifications regarding networking.
1. ARP must be sent and replied
Your camera has IP interface, which means it must handle ARP requests fine without any doubts. ARP is essential part of communicating via IP, camera without ARP support makes no sense. And ARP isn't a result of converting UDP, it's a preliminary step before sending actual UDP datagram. Once ARP reply is discovered with destination MAC-address, UDP packet is sent to that destination. The issue you see isn't about hardcoding MAC to avoid ARP.
2. Your code looks fine
Compiled it locally with minor corrections (missing #include <unistd.h> header with close() declaration), tested on several targets, client works as expected.
3. Something is wrong with your network topology
You are sending message from 192.168.1.70 to 10.20.1.2, which is weird. 192.168.0.0/24 and 10.0.0.0/8 are private IP addresses from different ranges, so they normally can't reach each other without black magic (like NAT traversal). And, what is much weirder, during your attempt ARP request is sent to strange destination. Let me illustrate different cases:
if both devices are in same subnet (e.g. 192.168.1.70 sends to 192.168.1.71), then message is sent directly, so client asks "who has 192.168.1.71" in ARP request.
if devices are in different subnets (e.g. 192.168.1.70 sends to 8.8.8.8), then message is sent through gateway, thus ARP request reads "who has 192.168.1.1" or whatever your gateway address is. Gateway MAC may be already in cache, in which case ARP isn't sent at all.
in your case subnets are obviously different, but ARP is asking about direct destination address rather than gateway MAC address.
It's a shot in the dark, but probably you have two network interfaces on your PC, one connected to 192.168.0.0 subnet, the other to 10.0.0.0 and ARP request is sent from both. If you sniff the wrong interface, you see weird ARP request and don't see UDP, which is actually sent after it. By the way, seeing single arp request is also confusing, because it should be repeated several times if noone answers.
Anyway, you need to check network topology and/or simplify it. Remove unnecessary network interfaces, configure PC and camera to be on the same subnet connected to the same switch/router and investigate further.
I have the situation where a service running in the background makes itself available for ASCII-based commands via a socket (SOCK_DGRAM) placed on the file system. I am able to successfully send commands to this interface but cannot receive any response generated by the background service.
As I understand it, the reason I am not receiving the service's response is because the underlying IPC is not technically between two processes, but is rather between to addresses. As such, it is necessary to bind my endpoint to a particular address location so the service knows were to send its response. However, the problem is that I do not want to pollute the directory space with too many additional socket files.
That is to say, I can make this work by simply doing something like:
struct sockaddr_un local;
int len;
s = socket(AF_UNIX, SOCK_DGRAM, 0);
local.sun_family = AF_UNIX;
strcpy(local.sun_path, "/path/to/some/dir/mySocketFile");
len = strlen(local.sun_path) + sizeof(local.sun_family);
bind(s, (struct sockaddr *)&local, len);
//Send commands to control interface of background service
And all is well, because by binding to mySocketFile the service has an address to which is will respond.
In short, is there a way to communicate to the service through its available socket interface and receive the response without binding the local endpoint such that it creates another socket-type file on the file system? i.e. some kind of a nameless socket, of sorts?
Of course, if anyone spots any misconceptions or misunderstandings in my logic please point them out.
If the client does not bind its socket to an filesystem address, it still has a notional address assigned by the system (which may exist in the filesystem in /tmp somewhere, or may not exist in the filesystem at all, depends on the OS). The server can get this address by using the recvfrom(2) call to receive the incoming packets from clients -- this call takes additional sockaddr * and socklen_t * arguments that it fills in with the client socket address. You then use sendto(2) to send the reply back to the client.
I want to connect clients to a server using ZeroMQ (java bindings, jzmq), but I need the TCP information badly, for example the TCP/IP address of a client request! The problem is, for being able to announce a service in the network I need to grab the TCP address of a request to be able to redirect clients to that service. The broker is a central "service registry" in that case. However, having ZeroMQ services on both sides, I do not see an option to retrieve that information.
What I do now, is to establish a dummy connection using a standard socket to the broker, after the connection is established I grab the IP address used for this connection and close the connection again. The IP address which has been retrieved is now being used for binding on it using a ZeroMQ socket on a random port.
I think this solution is the ugliest solution ever possible, so: What is a better solution to this problem?
Greetings.
0MQ doesn't provide the address of peers, for a number of reasons. It's also not that useful since what you really want is the endpoint to receive connections on, not the address the connection was made on.
What I usually do, and it's elegant enough, is pass bind a service to an ephemeral port, get a full connection endpoint ("tcp://ipaddress:port") and send that string in some way, either broadcast to peers, to a central registry, etc. along with my service name. Then, peers who want to connect back can take the service name, look up to find my endpoint, and connect back to me.
In ZMQ 4.x, you may get the string property "Peer-Address" or the "Identity" property. http://api.zeromq.org/4-2:zmq-msg-gets
The Identity is set in the other peer before connect(). http://api.zeromq.org/4-2:zmq-setsockopt#toc20
For example,
const char *identityString = "identity";
zmq::context_t context(1);
zmq::socket_t socket(context, ZMQ_REQ);
socket.setsockopt(ZMQ_IDENTITY, identityString, strlen(identityString));
socket.connect("tcp://127.0.0.1:5555");
Then the other side:
while(1)
{
zmq::message_t request;
if (socket.recv(&request, ZMQ_NOBLOCK))
{
const char* identity = request.gets("Identity");
const char* peerAddress = request.gets("Peer-Address");
printf("Received from %s %s\n", peerAddress, identity);
break;
}
}
I'm using CppZmq btw, you should be able to find the relevant calls easily.
Digging deeper into the libzmq code, I discovered that the library attaches to every message instance the file descriptor that it was received on.
This worked for me
int sockfd = zmq_msg_get(&msg, ZMQ_SRCFD);
sockaddr_in addr;
socklen_t asize = sizeof(addr);
getpeername(sockfd, (sockaddr*)&addr, &asize);
std::cout << inet_ntoa(addr.sin_addr) << ":" << addr.sin_port << std::endl;
Note that the FDs can and will be reused by other connections.
I'm working with version 4.2.1 of the api using the CZMQ binding and I found a solution for my case (ZMQ_STREAM). It works by setting an id before connecting.
The relevant socket option is "ZMQ_CONNECT_RID".
ZMQ api via zmq_setsockopt()
CZMQ api via zsock_set_connect_rid()
Some codes with redacted redacted ips.
const char endpoint1[] = "tcp://1.2.3.4:12345"
const char endpoint2[] = "tcp://5.6.7.8:12345"
zsock_t *stream = zsock_new(ZMQ_STREAM);
zsock_set_connect_rid(stream, endpoint1);
zsock_connect(stream, endpoint1);
zsock_set_connect_rid(stream, endpoint2);
zsock_connect(stream, endpoint2);
Then I get those 2 messages if there is a connection. First frame is the id and second frame is empty on connect/disconnect for ZMQ_STREAM sockets.
[Message1]
[019] tcp://1.2.3.4:12345
[000]
[Message2]
[019] tcp://5.6.7.8:12345
[000]
Another option is to use the zmq_socket_monitor() or czmq zmonitor. It was one of my first solution but I was looking for something lighter. I was able the get the endpoint that way without setting the id directly on the socket.
The zmonitor zactor make it possible to subscribe to socket events and then it sends a message with 3 frames:
[009] CONNECTED
[002] 14
[021] tcp://127.0.0.1:33445