4 ms·

The values of the FOSS community are pro Good. If you do something shady, and they help you save face with a gentle touch like this, you're lucky.

Dylan16807 3y ago
It's a shady workaround to the kernel itself being shady with which symbols get marked GPL.
- mort96 3y ago
  Marking symbols as GPL is not shady.
  - Dylan16807 3y ago
    For the symbols where use of them does not imply that the code is derivative of the kernel, it is shady. See also kernel_fpu_begin
    - mort96 3y ago
      Code which links against the kernel is derivative as far as the GPL is concerned.
      - deleted 3y ago
        [deleted]
      - Dylan16807 3y ago
        That's obviously not true when you consider code that was ported from other architectures. Unless you mean specifically how the GPL is written, but the GPL doesn't get to decide what's derivative, copyright does. And if that was solidly true, why isn't every symbol marked GPL?
        tristan957 3y ago
        Doesn't the existence of the LGPL negate your argument? It is the GPL with a linking exception, colloquially.
        Dylan16807 3y ago
        LGPL sets out better, clearer boundaries. And it allows some things that might arguably be derived code, while the GPL would hedge much closer to that line. So it has a use. Also FSF tends to overstate the importance of linking.
        mort96 3y ago
        I don't understand what porting from other architectures has to do with anything. If Linux has some code for amd64, and then Linux developers port that code to aarch64, why would the aarch64 code be different from a licensing perspective than the amd64 code? I don't know why they have non-GPL symbols. My understanding is that they could have marked every symbol as GPL. There's presumably some desire to allow limited closed-source drivers. You'd have to ask someone involved or find some writings from someone who knows. Maybe Google can help.
        Dylan16807 3y ago
        I'm taking about a kernel module being ported from something other than Linux. With minimal changes. Changing some entry points won't make it a derived work.

Dylan16807 3y ago

It's a shady workaround to the kernel itself being shady with which symbols get marked GPL.

mort96 3y ago
Marking symbols as GPL is not shady.
- Dylan16807 3y ago
  For the symbols where use of them does not imply that the code is derivative of the kernel, it is shady. See also kernel_fpu_begin
  - mort96 3y ago
    Code which links against the kernel is derivative as far as the GPL is concerned.
    - deleted 3y ago
      [deleted]
    - Dylan16807 3y ago
      That's obviously not true when you consider code that was ported from other architectures. Unless you mean specifically how the GPL is written, but the GPL doesn't get to decide what's derivative, copyright does. And if that was solidly true, why isn't every symbol marked GPL?
      - tristan957 3y ago
        Doesn't the existence of the LGPL negate your argument? It is the GPL with a linking exception, colloquially.
        Dylan16807 3y ago
        LGPL sets out better, clearer boundaries. And it allows some things that might arguably be derived code, while the GPL would hedge much closer to that line. So it has a use. Also FSF tends to overstate the importance of linking.
      - mort96 3y ago
        I don't understand what porting from other architectures has to do with anything. If Linux has some code for amd64, and then Linux developers port that code to aarch64, why would the aarch64 code be different from a licensing perspective than the amd64 code? I don't know why they have non-GPL symbols. My understanding is that they could have marked every symbol as GPL. There's presumably some desire to allow limited closed-source drivers. You'd have to ask someone involved or find some writings from someone who knows. Maybe Google can help.
        Dylan16807 3y ago
        I'm taking about a kernel module being ported from something other than Linux. With minimal changes. Changing some entry points won't make it a derived work.