3 ms·

The issue here is this appears likely to have been created by disassembly rather than pure study of behavior. Is that legal in Europe?

kwhitefoot 10y ago
Probably. No DMCA here.
- ntauthority 10y ago
  Directive 2009/24/EC has very similar implications, however.
charlesdm 10y ago
I'm not entirely sure of that. If you would have two teams though, it would. Team A creates a spec of the protocol, and Team B implements a connection lib based on the spec created by Team A. Europe tends to be pretty good through, copyright wise.
- PieterH 10y ago
  Yes, clean-room reimplementations are legal everywhere. The problem with this code is it's not clean room. This leaves commercial users open to serious accusations of copyright infringement (as in, criminal prosecution.) See my comments in the thread. IANAL and my analysis is based on the README of the project.
PieterH 10y ago
Disassembly by a team that results in new code written by the same team is not legal in any country that enforces copyrights. That includes the EU. Possibly excludes Russia and probably excludes China unless the owner is a large Chinese company. In fact I'd be surprised if this code isn't taken off GitHub by a DCMA from Microsoft, as soon as it hits their lawyers' radars.
- poizan42 10y ago
  > Disassembly by a team that results in new code written by the same team is not legal in any country that enforces copyrights. It's not illegal in itself. But you are easily opening yourself up to claims that the code you have written has been "tainted" by the disassembled code you have read and has therefore become a derivative work, which is going to be very to defend hard against. Also note that even if it is decided that you have made a derivative work you might still successfully launch a fair use defense on the grounds on interoperability, see e.g. Sega v. Accolade[1]. That said, doing a proper clean-room reverse engineering with two teams kept separate is a much safer approach. [1]: https://en.wikipedia.org/wiki/Sega_v._Accolade https://en.wikipedia.org/wiki/Sega_v._Accolade
  - PieterH 10y ago
    Thanks for correcting me.

kwhitefoot 10y ago

Probably. No DMCA here.

ntauthority 10y ago
Directive 2009/24/EC has very similar implications, however.

charlesdm 10y ago

I'm not entirely sure of that. If you would have two teams though, it would. Team A creates a spec of the protocol, and Team B implements a connection lib based on the spec created by Team A. Europe tends to be pretty good through, copyright wise.

PieterH 10y ago
Yes, clean-room reimplementations are legal everywhere. The problem with this code is it's not clean room. This leaves commercial users open to serious accusations of copyright infringement (as in, criminal prosecution.) See my comments in the thread. IANAL and my analysis is based on the README of the project.

PieterH 10y ago

Disassembly by a team that results in new code written by the same team is not legal in any country that enforces copyrights. That includes the EU. Possibly excludes Russia and probably excludes China unless the owner is a large Chinese company. In fact I'd be surprised if this code isn't taken off GitHub by a DCMA from Microsoft, as soon as it hits their lawyers' radars.

poizan42 10y ago
> Disassembly by a team that results in new code written by the same team is not legal in any country that enforces copyrights. It's not illegal in itself. But you are easily opening yourself up to claims that the code you have written has been "tainted" by the disassembled code you have read and has therefore become a derivative work, which is going to be very to defend hard against. Also note that even if it is decided that you have made a derivative work you might still successfully launch a fair use defense on the grounds on interoperability, see e.g. Sega v. Accolade[1]. That said, doing a proper clean-room reverse engineering with two teams kept separate is a much safer approach. [1]: https://en.wikipedia.org/wiki/Sega_v._Accolade https://en.wikipedia.org/wiki/Sega_v._Accolade
- PieterH 10y ago
  Thanks for correcting me.