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All of this is irrelevant. The GPL stops at the process boundary and that's exactly what these compilers do. They provide the modifications to GCC itself under
by the_mitsuhiko 10y ago
All of this is irrelevant. The GPL stops at the process boundary and that's exactly what these compilers do. They provide the modifications to GCC itself under the GPL and then have separate executables that are proprietary that feed things in and out of GCC.
- nitrogen 10y agoIt's still pretty clearly a violation of the intentions of the gcc developers, and something that only exists to wrap one product seems very suspiciously like a derivative work. I'd be curious to see this tested in court.
- the_mitsuhiko 10y agoBut the FSF would not be curious to test this in court because they know they lose. I had this conversation many years ago with people that wrote extactly this type of software and the general consensus was that the FSF would stay away from probing the GPL there because there is angood chance that the ruling would set a precedent they don't want. The FUD surrounding the license in that regard is preferrable. In general it's 2016 and no longer relevant. The world moved on to LLVM and these GCC using toolchains are on their way out for good.
- lmm 10y ago> The GPL stops at the process boundary No it doesn't, and if it did that would only mean that processes that form derivative works of GPLed programs were unlicensed, and therefore a violation of copyright.
- dragonwriter 10y agoThe text of the GPL (nor the FSF FAQ) does not explicitly support the "stops at process boundaries" interpretation, though a form of that's been a popular interpretation since at least the GPLv2 period.