Get Multicast Dest from Python Socket Recv
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For my project, I need some way to know which multicast group a packet was originally sent to when it's received by my client. I've considered maintaining a map of sockets to multicast groups and identifying them this way, but surely there's some way to get the address from the datagram?
To listen, I'm currently using:
# Initialise socket for IPv6 datagrams
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, socket.IPPROTO_UDP)
# Allows address to be reused
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
# Binds to all interfaces on the given port
sock.bind(('', 8080))
# Allow messages from this socket to loop back for development
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
# Construct message for joining multicast group
mreq = struct.pack("16s15s".encode('utf-8'), socket.inet_pton(socket.AF_INET6, "ff02::abcd:1"), (chr(0) * 16).encode('utf-8'))
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_JOIN_GROUP, mreq)
data, addr = sock.recvfrom(1024)
and to send:
# Create ipv6 datagram socket
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
# Allow own messages to be sent back (for local testing)
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
sock.sendto("hello world".encode('utf-8'), ("ff02::abcd:1", 8080))
Which works, but the address when received is that of the sending machine. How can I see the multicast group it was sent to?
Thanks!
python python-3.x sockets udp multicast
add a comment |
For my project, I need some way to know which multicast group a packet was originally sent to when it's received by my client. I've considered maintaining a map of sockets to multicast groups and identifying them this way, but surely there's some way to get the address from the datagram?
To listen, I'm currently using:
# Initialise socket for IPv6 datagrams
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, socket.IPPROTO_UDP)
# Allows address to be reused
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
# Binds to all interfaces on the given port
sock.bind(('', 8080))
# Allow messages from this socket to loop back for development
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
# Construct message for joining multicast group
mreq = struct.pack("16s15s".encode('utf-8'), socket.inet_pton(socket.AF_INET6, "ff02::abcd:1"), (chr(0) * 16).encode('utf-8'))
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_JOIN_GROUP, mreq)
data, addr = sock.recvfrom(1024)
and to send:
# Create ipv6 datagram socket
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
# Allow own messages to be sent back (for local testing)
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
sock.sendto("hello world".encode('utf-8'), ("ff02::abcd:1", 8080))
Which works, but the address when received is that of the sending machine. How can I see the multicast group it was sent to?
Thanks!
python python-3.x sockets udp multicast
add a comment |
For my project, I need some way to know which multicast group a packet was originally sent to when it's received by my client. I've considered maintaining a map of sockets to multicast groups and identifying them this way, but surely there's some way to get the address from the datagram?
To listen, I'm currently using:
# Initialise socket for IPv6 datagrams
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, socket.IPPROTO_UDP)
# Allows address to be reused
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
# Binds to all interfaces on the given port
sock.bind(('', 8080))
# Allow messages from this socket to loop back for development
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
# Construct message for joining multicast group
mreq = struct.pack("16s15s".encode('utf-8'), socket.inet_pton(socket.AF_INET6, "ff02::abcd:1"), (chr(0) * 16).encode('utf-8'))
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_JOIN_GROUP, mreq)
data, addr = sock.recvfrom(1024)
and to send:
# Create ipv6 datagram socket
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
# Allow own messages to be sent back (for local testing)
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
sock.sendto("hello world".encode('utf-8'), ("ff02::abcd:1", 8080))
Which works, but the address when received is that of the sending machine. How can I see the multicast group it was sent to?
Thanks!
python python-3.x sockets udp multicast
For my project, I need some way to know which multicast group a packet was originally sent to when it's received by my client. I've considered maintaining a map of sockets to multicast groups and identifying them this way, but surely there's some way to get the address from the datagram?
To listen, I'm currently using:
# Initialise socket for IPv6 datagrams
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, socket.IPPROTO_UDP)
# Allows address to be reused
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
# Binds to all interfaces on the given port
sock.bind(('', 8080))
# Allow messages from this socket to loop back for development
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
# Construct message for joining multicast group
mreq = struct.pack("16s15s".encode('utf-8'), socket.inet_pton(socket.AF_INET6, "ff02::abcd:1"), (chr(0) * 16).encode('utf-8'))
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_JOIN_GROUP, mreq)
data, addr = sock.recvfrom(1024)
and to send:
# Create ipv6 datagram socket
sock = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
# Allow own messages to be sent back (for local testing)
sock.setsockopt(socket.IPPROTO_IPV6, socket.IPV6_MULTICAST_LOOP, True)
sock.sendto("hello world".encode('utf-8'), ("ff02::abcd:1", 8080))
Which works, but the address when received is that of the sending machine. How can I see the multicast group it was sent to?
Thanks!
python python-3.x sockets udp multicast
python python-3.x sockets udp multicast
edited Nov 26 '18 at 20:00
Jordan Mackie
asked Nov 26 '18 at 19:38
Jordan MackieJordan Mackie
4261523
4261523
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1 Answer
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You only joined multicast group address "ff02::abcd:1". Therefore, any packet that is received on the socket must have been sent to that multicast address.
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You only joined multicast group address "ff02::abcd:1". Therefore, any packet that is received on the socket must have been sent to that multicast address.
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
add a comment |
You only joined multicast group address "ff02::abcd:1". Therefore, any packet that is received on the socket must have been sent to that multicast address.
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
add a comment |
You only joined multicast group address "ff02::abcd:1". Therefore, any packet that is received on the socket must have been sent to that multicast address.
You only joined multicast group address "ff02::abcd:1". Therefore, any packet that is received on the socket must have been sent to that multicast address.
answered Jan 8 at 15:07
Bruno RijsmanBruno Rijsman
1,70331936
1,70331936
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
add a comment |
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
Thanks Bruno. I understand that, I'm trying to solve the problem for one socket that joins multiple multicast groups though.
– Jordan Mackie
Jan 9 at 20:00
add a comment |
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