My software need to listen to 2 different ports at the same time, I need something like that :
Client.Connect("127.0.0.1", 8001);
Client.Connect("127.0.0.1", 8002);
Is it possible ?
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Alexus
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4Why wouldn't you just make two `TcpClient`s? – Glorin Oakenfoot Nov 19 '15 at 15:05
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1To the best of my knowledge this isn't possible with the same client object, but of those you can have as many as you need, as suggested by Glorin Oakenfoot. – 500 - Internal Server Error Nov 19 '15 at 15:07
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3Also, if you are *listening*, as in listening for connections, you should probably have a look at the `TcpListener` class. – Glorin Oakenfoot Nov 19 '15 at 15:08
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1... and if you want to listen on two separate ports, create two listeners... – Jon Skeet Nov 19 '15 at 15:12
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1Just use two separate TcpClients. It makes no sense to want to use the same one. – CathalMF Nov 19 '15 at 15:18
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Ok, thanks for your answers. – Alexus Nov 19 '15 at 15:21
1 Answers
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Not with a 1 connection, TCP Clients are connected individually so you need to make 2 port connection for multiple connectivity. For example, in Terminal you can't have multiple connection until you open a new window for second connection.
See Borealid answer on here, give you some clear view, but there is more to it.
Brief explanation:
For a stateful protocol (like TCP), a connection is identified by a 4-tuple consisting of source and destination ports and source and destination IP addresses. So, if two different machines connect to the same port on a third machine, there are two distinct connections because the source IPs differ. If the same machine (or two behind NAT or otherwise sharing the same IP address) connects twice to a single remote end, the connections are differentiated by source port (which is generally a random high-numbered port).
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Brian Nezhad
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1My software have two servers listening two devices, one server in TCP, the other in UDP mode. The UI application is listening to that two servers and displays the informations. I added another TcpClients and that seems to work. – Alexus Nov 19 '15 at 16:49