WebRTC IOS SDK API documentation

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Neha Yadav

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May 21, 2015, 5:30:16 PM5/21/15
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Hi,

I have compile the libraries for IOS webrtc as mentioned on webrtc official site.
Now I want to make a project that uses IOS SDK apis provided for webrtc. But I have not found any documentaion for this.
Every documenation is providedd with respect to browsers but not if we need to include webrtc support in voip app.

Can somebody provide me a API doc how to use ios sdk provided by google for webrtc.

Thanks
Neha

Christoffer Jansson

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May 25, 2015, 3:03:12 AM5/25/15
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Hi Neha,

You can find object C wrapper files here and some instructions in the parent folder README file here.

I hope that can be of some help.

/Chris

evv rajesh

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Jun 5, 2015, 3:52:54 AM6/5/15
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Hi ,

I am trying to compile webrtc for ios platform.
Can you just specify the process to compile.
Because, i am facing calledprocess error when working with fetch webrtc_ios.

Can you just help me out to fix this problem?

It will be helpfull.

Christoffer Jansson

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Jun 5, 2015, 3:55:06 AM6/5/15
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You can find instructions here. Follow them step by step, if something does not work, please attach the steps you did and the error output.

Siya Kunde

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Feb 1, 2017, 5:27:09 AM2/1/17
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Hey Christoffer,

The links you have mentioned are not working. Can you please point me to where I could find a proper documentation of iOS WebRTC APIs?

Thanks,
Siya Kunde

Henrik Kjellander

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Feb 10, 2017, 3:29:24 AM2/10/17
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Alvaro Gil

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Feb 10, 2017, 8:27:16 AM2/10/17
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Siya, as far as I know there is not such thing as IOS documentation, there are guides to build[0] the library and you can read the IOS wrapper headers that outline the IOS API[1] to understand what it has. However the simplest entry point to its API is the AppRTCMobile[2] demo app, you will want to check it out and see how it works.

Most important is to understand how WebRTC works no matter the client you are using cause his behavior and API's are very similar across platforms. I think you can get the big picture from one of the resources here[3].

Hope it helps!


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