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Osquery: Change license to Apache 2.0 and GPLv2

46 点作者 zwass超过 7 年前

2 条评论

jsmthrowaway超过 7 年前
Since Apache License v2.0 and GPLv2 are incompatible, I&#x27;d be curious to understand the implications of this license selection for a few scenarios in both directions. Say osquery wanted to include a GPLv2 library. Could it? How about an Apache License v2.0 library? ASF and FSF differ on the interpretation of the details, which would come into play here. Does the GPLv2&#x27;s requirement that it apply to the whole work[0] matter?<p>A more interesting thought: say I wanted to fork osquery. Do I select one of the licenses as part of my forking operation, since I, the user, am exercising my rights under one of the licenses (at my option) to derive the work? Would that mean I could fork osquery to Apache License v2.0, GPLv2, or both at my discretion? Given the requirements of GPLv2, am I even permitted to do that?<p>An even more interesting thought: osquery seems to <i>derive</i> GPLv2 code from the Netfilter project by patching it:<p><a href="https:&#x2F;&#x2F;github.com&#x2F;facebook&#x2F;osquery&#x2F;blob&#x2F;master&#x2F;tools&#x2F;provision&#x2F;formula&#x2F;libiptables.rb#L35-L50" rel="nofollow">https:&#x2F;&#x2F;github.com&#x2F;facebook&#x2F;osquery&#x2F;blob&#x2F;master&#x2F;tools&#x2F;provis...</a><p><a href="https:&#x2F;&#x2F;github.com&#x2F;facebook&#x2F;osquery&#x2F;blob&#x2F;master&#x2F;tools&#x2F;provision.sh#L100" rel="nofollow">https:&#x2F;&#x2F;github.com&#x2F;facebook&#x2F;osquery&#x2F;blob&#x2F;master&#x2F;tools&#x2F;provis...</a><p>Can the resulting product then be distributed Apache License v2.0, given the viral nature of GPLv2? I&#x27;m not clear how that dependency is used, but is Facebook even able to license this work as Apache License v2.0 at all?<p>This is a very confusing dual-license scenario, and something I&#x27;d never thought about with choices like Apache&#x2F;MIT. I feel it important to point out that I&#x27;m not attempting to introduce FUD or criticize Facebook; I&#x27;m genuinely curious.<p>[0]: <a href="https:&#x2F;&#x2F;opensource.stackexchange.com&#x2F;a&#x2F;1364" rel="nofollow">https:&#x2F;&#x2F;opensource.stackexchange.com&#x2F;a&#x2F;1364</a>
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kirillzubovsky超过 7 年前
What is the meaning of this?
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