CODEWITHSUNDEEP
javascriptwebrtcstun
Dlacreme
Dlacreme

Reputation: 206

How does RTCPeerConnection use the array of STUN/TURN servers?

We are doing some research regarding the behavior of Web RTC with the list of STUN/TURN servers provided. I cannot find any documentation so I am doing some tests but I hope someone can provide a clear explanation.

Following the documentation (https://developer.mozilla.org/en-US/docs/Web/API/RTCIceServer/urls) we can provide any number of servers.

To provide more context, we have a working WebRTC application using the Google Stun Servers (stun.l.google.com:19302) however we are migrating on our own STUN servers. We have an API that returns the list of STUN servers to use however based on the behavior we might provide a different list.

Thanks for your help

Upvotes: 2

Views: 531

Answers (1)

Sean DuBois
Sean DuBois

Reputation: 4242

But how does the RTCPeerConnection choose the server to use?

The RTCPeerConnection doesn't choose a server it choose a pair of ICE candidates. A ICE candidate is generated by contacting the servers.

Does it try the first one and if it fails try the second one until one works?

It contacts them all (this process is called gathering). Your WebRTC implementation may stop gathering when a connection is established.

Should all the servers be up and running or is the connection able to skip an unreachable server?

It is ok for a server to be down. Trickle ICE allows connectivity checks to proceed if all of them are not working.

If the first server is able to conclude the negociation, does it still try with the remaining servers?

Two WebRTC Agents don't communicate through a STUN server, this question has a little more nuance. In the connectivity chapter of WebRTC for the Curious check out how ICE goes through steps.

Does it simply change the list of candidates?

Yea! For each STUN server you may have another candidate. This depends on the behavior of your NAT. You could be behind a NAT configuration that only gives you one mapping. Unlikely but still possible!

Upvotes: 5

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