5 ms·

Is Android really free software? - Richard Stallman

27 points by Anilm3 15y ago

j_col 15y ago
From the article: > Thus, Android 3, apart from Linux, is non-free software, pure and simple. Couldn't agree more, happy to see Stallman call it: Android 3.x is clearly not open source.
- doki_pen 15y ago
  I don't think he cares about open source. He's saying it isn't free software.
ZeroGravitas 15y ago
Google has said it will never publish the source code of Android 3.0 (aside from Linux), even though executables have been released to the public. Android 3.1 source code is also being withheld. Thus, Android 3, apart from Linux, is non-free software, pure and simple. I was under the impression that this wouldn't be possible due to their use of git, and so while they aren't making a big deal of it, you could, if you so desired figure out at which point they branched for 3.0 or 3.x and create your own version, they're just not going to make it easy because they'd rather you use the newer code built on top of that. I assumed he was going to say it's not free software because they've not yet released the code above the linux level. If he's saying that it's only not considered free software by him because it will never be released, then I'd be interested in what he'd think if it's simply delayed and semi-obfuscated as part of a later version release. Also, is Replicant targetting the Galaxy S II? I thought I read that it had the best support for the hardware without binary of proprietary chunks, but I don't see any mention of it on their site.
- doki_pen 15y ago
  If effort has been made to obfuscate code, then it's not free software. Free software doesn't hinder your ability to share, run, modify and or share modifications. That's what makes it free.
  - ZeroGravitas 15y ago
    It's not the code that's hidden though, it's exactly which revisions correspond to the various releases. That's still not in the spirit of free software, but if later versions are better, freely available and almost entirely a superset then it becomes a bit of a pedantic distinction.

27 points by Anilm3 15y ago

j_col 15y ago

From the article: > Thus, Android 3, apart from Linux, is non-free software, pure and simple. Couldn't agree more, happy to see Stallman call it: Android 3.x is clearly not open source.

doki_pen 15y ago
I don't think he cares about open source. He's saying it isn't free software.

ZeroGravitas 15y ago

Google has said it will never publish the source code of Android 3.0 (aside from Linux), even though executables have been released to the public. Android 3.1 source code is also being withheld. Thus, Android 3, apart from Linux, is non-free software, pure and simple. I was under the impression that this wouldn't be possible due to their use of git, and so while they aren't making a big deal of it, you could, if you so desired figure out at which point they branched for 3.0 or 3.x and create your own version, they're just not going to make it easy because they'd rather you use the newer code built on top of that. I assumed he was going to say it's not free software because they've not yet released the code above the linux level. If he's saying that it's only not considered free software by him because it will never be released, then I'd be interested in what he'd think if it's simply delayed and semi-obfuscated as part of a later version release. Also, is Replicant targetting the Galaxy S II? I thought I read that it had the best support for the hardware without binary of proprietary chunks, but I don't see any mention of it on their site.

doki_pen 15y ago
If effort has been made to obfuscate code, then it's not free software. Free software doesn't hinder your ability to share, run, modify and or share modifications. That's what makes it free.
- ZeroGravitas 15y ago
  It's not the code that's hidden though, it's exactly which revisions correspond to the various releases. That's still not in the spirit of free software, but if later versions are better, freely available and almost entirely a superset then it becomes a bit of a pedantic distinction.