When creating a peer to peer audio connection using webRTC, the STUN server we use will return the public IP if a user is behind a router. Now in the ICE objects, I can see that the rport is always something between 50000 and up.
Is there a way to use a specific port so that the user does not have to open all those ports?
Is there a way to use a specific port so that the user does not have to open all those ports?
I think you have a misunderstanding. The whole point of STUN and ICE (including its WebRTC derivative) exists to avoid anyone having to open a port on their NAT. Instead, STUN and ICE dynamically open the port.
Here's how it works (in a really brief description).
Client opens a socket on a random port (e.g. 50001)
Contacts STUN server using that socket to discover the external IP:port mapping for this socket. (e.g. 192.168.1.2:50001 maps to 1.2.3.4:50001). Ports don't necessarily have to match between internal and external addresses, but they usually do, so I'll keep with that for this example.
Through an external mechanism (SIP, XMPP, Jingle, cups with strings), the candidate address list of both nodes are exchanged. This includes all known internal and external addresses collected (e.g. 192.168.1.2:50001 and 1.2.3.4:50001).
Using the same socket opened in step 1, both sides send (STUN) messages (UDP packets) directly between each other. The first pair of messages may be blocked by the router/firewall. But because one side initiated an outbound packet to the remote address, subsequent packets from that address are allowed back in. This is called the "hole punching step". Hence, the port is dynamically open without the router needing any specific configuration.
Hope this helps.
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