[evmar]

I noticed math_emulate.c comes from Linux (it even has a " * (C) 1991 Linus Torvalds" bit on it). I was wondering what the license on that code is. It looks like Linux adopted GPL in 1992 so maybe this copy predates that, but it was under some other non-BSD license before that.

[einr]

The full license for Linux prior to 0.12 was:

  This kernel is (C) 1991 Linus Torvalds, but all or part of it may be
  redistributed provided you do the following:

  - Full source must be available (and free), if not with the
    distribution then at least on asking for it.

  - Copyright notices must be intact. (In fact, if you distribute
    only parts of it you may have to add copyrights, as there aren't
    (C)'s in all files.) Small partial excerpts may be copied
    without bothering with copyrights.

  - You may not distibute this for a fee, not even "handling"
    costs.

This is clearly written by someone who has no business writing software licenses ;) but does not appear to be incompatible with the BSD license and in fact, the code in question originates from 386BSD (https://github.com/386bsd/386bsd/blob/2.0/usr/src/kernel/mat...) and made it from there into the NetBSD mainline.

[nikanj]

I don't think the BSD license is compatible with "you may not distribute this for a fee"

[LeFantome]

The GPL has wrongly taught us to focus on “compatibility”. Compatibility is pretty much exclusively a copyleft issue.

There is no reason that you cannot use this code with this license in a larger BSD work. It is “compatible” in that sense.

This specific code has additional restrictions (not charging). That does not add any restrictions to the rest of the code though.

So, if you are charging, you are in violation of the license just for this code snippet. Linus Torvalds, the copyright holder, could try to enforce the license against you. Since he gives it away, no financial damages. Which means the remedy would be that you would not be able to use this code anymore (but could still use the rest of BSD).

I do not expect Linus to pursue enforcement on this one.

It would be a very difficult case to win anyway as you would have to prove that people were being charged specifically for the Linus code and not for other code covered by BSD (which allows charging).

I would argue that this license has not even been violated, unless somebody has put a price tag on this specific code.

[einr]

True! And if so, that license has clearly been broken many times by everyone selling 386BSD, NetBSD and Linux <0.12 on CD-ROMs etc ;)

Then again -- and IANAL -- the license is worded so vaguely that I doubt any of it is enforcible. "You may not distibute this for a fee" -- what is "this"? Is it the entire kernel or does it apply to small excerpts of it? Because apparently "small partial excerpts may be copied without bothering with copyrights". But do you mean copyright attribution or are you rescinding your copyright entirely if I only copy "small partial excerpts"? But what is a small partial excerpt? And so on and so forth...

[LeFantome]

“this” is the code that Linus licensed as he did. Only that code. If you use a snippet, that snippet is governed by the license. The license does not magically extend to other code or even to any modifications that have been made since. This is not the GPL.

[wkat4242]

I think nikanj means that this is not a stipulation of the BSD license at all, that's why it's not compatible with what Linus made there.

[LeFantome]

But the Linus license has no bearing on the rest of the code base at all.

The entire concept of “compatibility” is an artifact of copyleft. In the rest of the license universe, each piece of code is covered by its own license and it does not matter what licenses other code uses.

This license does not apply to the rest of BSD. The BSD license does not apply to this code.

[yjftsjthsd-h]

Why not? BSD style licenses generally just impose constraints around liability and attribution; I can't see any reason why that wouldn't be compatible with a separate constraint on charging money. IANAL, though, so take with grain of salt.