JerGrant / QiskitQuantumStatistics

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Add a license to the repository #1

Open cbjuan opened 3 weeks ago

cbjuan commented 3 weeks ago

I recently came across your project QiskitQuantumStatistics and I noticed that the repository github.com/JerGrant/QiskitQuantumStatistics currently does not have a license specified.

Why Adding a License is Important?

As GitHub documents

Public repositories on GitHub are often used to share open source software. For your repository to truly be open source, you'll need to license it so that others are free to use, change, and distribute the software.

Would be possible to add a valid license to this repository? Probably something really OpenSource such as the Apache 2.0 license would fit really well.

Thanks!

JerGrant commented 5 days ago

Will do! Thanks for the notice, a definite oversight on my part.

On Wed, Sep 4, 2024, 5:02 AM Juan Cruz-Benito @.***> wrote:

I recently came across your project QiskitQuantumStatistics and I noticed that the repository github.com/JerGrant/QiskitQuantumStatistics currently does not have a license specified.

Why Adding a License is Important?

As GitHub documents https://docs.github.com/en/repositories/managing-your-repositorys-settings-and-features/customizing-your-repository/licensing-a-repository

Public repositories on GitHub are often used to share open source software. For your repository to truly be open source, you'll need to license it so that others are free to use, change, and distribute the software.

Would be possible to add a valid license to this repository? Probably something really OpenSource such as the Apache 2.0 license https://www.apache.org/licenses/LICENSE-2.0 would fit really well.

Thanks!

— Reply to this email directly, view it on GitHub https://github.com/JerGrant/QiskitQuantumStatistics/issues/1, or unsubscribe https://github.com/notifications/unsubscribe-auth/AZBGWQRBHACOLFCK6PA7WWDZU3SETAVCNFSM6AAAAABNUBZATWVHI2DSMVQWIX3LMV43ASLTON2WKOZSGUYDKMBRGI3DGOI . You are receiving this because you are subscribed to this thread.Message ID: @.***>