Re: [Linphone-developers] With the announcement from Google about 64 bit support for apps on the Play Store, I am struggling to get the 4.0.1 SDK working with 64 bit.

[Sylvain Berfini]

Hi,

The repository you are using is the oldest one, that is why you don't 
see the latest versions.

You can find the 4.2 in 
http://linphone.org/releases/maven_repository/org/linphone/linphone-sdk-android/ 
repository but it is now deprecated.
The all new repository that mixes snapshots & releases base on the 
semantic versioning name is available at: 
https://linphone.org/maven_repository/org/linphone/linphone-sdk-android/

It contains the latest 4.2 releases, the 4.3 alpha & beta versions and 
soon the 4.3 release.

Cheers,

Le 05/12/2019 à 16:00, iv gregory a écrit :
> Dear Linphone community, please, help...
>
> With the announcement from Google about 64 bit support for apps on the Play 
> Store, I am struggling to get the 4.0.1 SDK working with 64 bit.
>
> Here is the link to the Google explanation announcement itself:
> https://developer.android.com/distribute/best-practices/develop/64-bit
>
> On the link bellow I have found the similar problem description which I faced 
> but with other library type (linphone):
> https://lists.nongnu.org/archive/html/linphone-developers/2019-07/msg00025.html
>
> The solution which they recommend is to use the following lib (also linphone) :
> https://www.linphone.org/releases/maven_repository/org/linphone/linphone-sdk-android/4.2/linphone-sdk-android-4.2.aar
>
>
> But In my case I use the following lib and there is no 4.2 version available 
> (liblinphone) :
> https://www.linphone.org/releases/maven_repository/org/linphone/liblinphone-sdk/4.0.1/liblinphone-sdk-4.0.1.aar
>
> List of available liblinphone libraries (no 4.2 version available) :
> https://www.linphone.org/releases/maven_repository/org/linphone/liblinphone-sdk/
>
> Please, help to find the solution.
>
> Do they suppose to release liblinphone-sdk-4.2.aar ?
>
> Thank you everyone for any help.
>
> Best regards and best whishes.
>
> Gregory.
>
>
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