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Using NVidia FxAA ​​in my code: What is the licensing model?

This implementation of FxAA ​​from NVidia has occurred.

http://developer.download.nvidia.com/assets/gamedev/files/sdk/11/FXAA_WhitePaper.pdf http://timothylottes.blogspot.in/2011/12/fxaa-40-stills-and-features.html

The source code as such does not have a license header, and I cannot find any licensing information. Is it possible to integrate this with commercially licensed software?

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2 answers

This is a public domain, obviously:

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For what it's worth, updates / derivatives from FXAA 3.11 are now also licensed under NVIDIA GameWorks OpenGL Examples :

  • FXAA 3.11
  • License , the essence of which is summarized in this passage:

    License: in accordance with the terms of this Agreement, NVIDIA hereby grants the Developer an irrevocable non-exclusive license to own and use materials. The following terms apply to the indicated type of material:

    Source code: the developer has the right to modify and create derivatives works with source code. The Developer shall (“Derivatives”), he creates the source code, provided that the Developer uses the Materials in accordance with the terms of this Agreement. Developer may distribute Derivatives provided that all notices and trademarks of NVIDIA are used and Derivatives include the following statement: "This software contains source code provided by NVIDIA Corporation."

This means that it requires attribution, but is still significantly more permissive than many other software licenses in relation to commercial use of the software.

Regarding the search for the initial disclosure of how the code will be licensed, I had a little trawl through the Internet archive and (possibly the suspect) Archive.is . There did not seem to be any mention of the intent of licensing, although other people's blog posts indicate this in the public domain:

It seems sigman's answer is about as good as it is in terms of support for using the public domain as code (outside the NVIDIA license linked above). Or outside the appeal to Timothy Lotte himself (again) .;)

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Source: https://habr.com/ru/post/924022/

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