P
Pratap Das
Hi All,
I have a question regarding an IPv6 datagram(UDP) server using
winsock2.
In my sample program , I am open a socket, bind it and then wait on a
recvfrom for data from clients. I can see in the output of "netstat
-an" that it is listening on the designated port.
But from a different machine the client is not able to communicate to
the server. Using a network sniffer I am seeing that the packet from
the client is reaching the server machine but not the server itself,
it is still waiting on the recvfrom.
Now, in the server if I make a call to sendto to the client before
waiting on recvfrom, I am able to send data from the client from the
next packet onwards. So looks like I have to make a call to sendto
before recvfrom to actually start receiving some data on the socket.
I have the IPv4 version of the program which works fine. Both are
examples from the MSDN library. If anyone can provide any pointers
reagrding the above behaviour in the case of IPv6 it will be really
helpful.
Thanks for all the help.
Regards,
--Das
I have a question regarding an IPv6 datagram(UDP) server using
winsock2.
In my sample program , I am open a socket, bind it and then wait on a
recvfrom for data from clients. I can see in the output of "netstat
-an" that it is listening on the designated port.
But from a different machine the client is not able to communicate to
the server. Using a network sniffer I am seeing that the packet from
the client is reaching the server machine but not the server itself,
it is still waiting on the recvfrom.
Now, in the server if I make a call to sendto to the client before
waiting on recvfrom, I am able to send data from the client from the
next packet onwards. So looks like I have to make a call to sendto
before recvfrom to actually start receiving some data on the socket.
I have the IPv4 version of the program which works fine. Both are
examples from the MSDN library. If anyone can provide any pointers
reagrding the above behaviour in the case of IPv6 it will be really
helpful.
Thanks for all the help.
Regards,
--Das