Linked by David Adams on Wed 10th Aug 2011 17:12 UTC, submitted by R_T_F_M
Apple The lure of shiny toys has helped Apple's BSD-based Mac OS X operating system overtake Linux to become the operating system that is the second most used by developers, according to Evans Data.
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Ready to Rumble?
by churlish_Helmut on Wed 10th Aug 2011 20:56 UTC
churlish_Helmut
Member since:
2010-04-12

In the left corner: We have the Linux User, denying that OSX is still a BSD.

In the right corner: The BSD Users, claiming that with OSX makes BSD more popular.

Lets get ready to rumble ;)

No, Serious.
I think OSX is definite part of the BSD Family. I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.

Reply Score: 4

RE: Ready to Rumble?
by Vanders on Wed 10th Aug 2011 21:07 in reply to "Ready to Rumble?"
Vanders Member since:
2005-07-06

I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.

By the logic that Android is Linux, OS X would be Mach.

I'd love to see an analysis of OS X to see just how much BSD derived code there is in there compared to code from other sources.

Reply Parent Score: 7

RE[2]: Ready to Rumble?
by itanic on Thu 11th Aug 2011 08:15 in reply to "RE: Ready to Rumble?"
itanic Member since:
2008-08-03

Considering Mach was derived from BSD, and then parts of it like the virtual memory management were integrated back into BSD, it doesn't make much sense to argue that something's not BSD because it's Mach.

Reply Parent Score: 2

RE[2]: Ready to Rumble?
by demetrioussharpe on Sat 13th Aug 2011 05:20 in reply to "RE: Ready to Rumble?"
demetrioussharpe Member since:
2009-01-09

"I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.

By the logic that Android is Linux, OS X would be Mach.

I'd love to see an analysis of OS X to see just how much BSD derived code there is in there compared to code from other sources.
"

Mach doesn't really have a userland. In order to get a working Mach system up & running, the normal approach is to use a BSD userland. So, no matter how you look at it, MacOS X's architecture is consistent with various parts of the BSD tree.

Reply Parent Score: 1

RE: Ready to Rumble?
by Gusar on Wed 10th Aug 2011 21:16 in reply to "Ready to Rumble?"
Gusar Member since:
2010-07-16

when you call "android" a linux, then OSX is a BSD.

I don't call Android a Linux (where Linux is defined as Linux distro the likes of Fedora, Ubuntu, etc). Sure it uses the Linux kernel and some userspace utils are common, but otherwise the Android userspace is very different from traditional Linux distros. Particularly in the area that matters most - the application framework.
Same with BSD and OSX - the application framework on OSX (Cocoa and the Core* stuff) is very different from the BSDs, which are more similar to Linux distros in that regard actually.

Edited 2011-08-10 21:18 UTC

Reply Parent Score: 4

RE[2]: Ready to Rumble?
by demetrioussharpe on Sat 13th Aug 2011 05:24 in reply to "RE: Ready to Rumble?"
demetrioussharpe Member since:
2009-01-09

"when you call "android" a linux, then OSX is a BSD.

I don't call Android a Linux (where Linux is defined as Linux distro the likes of Fedora, Ubuntu, etc). Sure it uses the Linux kernel and some userspace utils are common, but otherwise the Android userspace is very different from traditional Linux distros. Particularly in the area that matters most - the application framework.
Same with BSD and OSX - the application framework on OSX (Cocoa and the Core* stuff) is very different from the BSDs, which are more similar to Linux distros in that regard actually.
"

Congratulations, you've just proved that the userland is a horrible way to determine what family an OS is in. Afterall, OS/2 isn't Windows 3.1, but you can run programs written for Windows 3.1 on it. You can run Linux programs on FreeBSD, that doesn't give Linux & FreeBSD family ties. The fact of the matter is that BSD & Mach DNA (figuratively) are in the core of MacOS X. Really, that's all that matters. In matters of genealogy, it's not what you have now, it's what you started with.

Reply Parent Score: 1

RE: Ready to Rumble?
by andydread on Wed 10th Aug 2011 22:34 in reply to "Ready to Rumble?"
andydread Member since:
2009-02-02

In the left corner: We have the Linux User, denying that OSX is still a BSD.

In the right corner: The BSD Users, claiming that with OSX makes BSD more popular.

Lets get ready to rumble ;)

No, Serious.
I think OSX is definite part of the BSD Family. I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.


As a Linux and Windows user i have to agree with them on this OSX is basically BSD or one can say OSX is powered by BSD. The UI is proprietary but the OS kernel and userland is mostly based on BSD.

Reply Parent Score: 1

RE: Ready to Rumble?
by brynet on Wed 10th Aug 2011 23:10 in reply to "Ready to Rumble?"
brynet Member since:
2010-03-02

Android is a Java OS running in a Google authored JavaVM running on top of a Linux kernel.

There have been musings of porting it to other kernels, I think it has even been partially ported to QNX.

So it's not really the same thing, Android isn't native while an OS X application using Apple's proprietary libraries/kits can still make use of the BSD environment without issues.

Even iOS's SDK apparently exposed a fair amount of the BSD environment to applications, with some limitations.

Edited 2011-08-10 23:10 UTC

Reply Parent Score: 2

RE[2]: Ready to Rumble?
by ebasconp on Thu 11th Aug 2011 04:19 in reply to "RE: Ready to Rumble?"
ebasconp Member since:
2006-05-09

I have mixed "feelings" on all this BSD topic, because the Mac and all the BSDs share the same codebase; a lot of Mac low level software has been ported to the BSDs and Linux and a lot of UNIX apps run with no problems in the BSD layer of Mac OS X.

But... you can use KDE, Gnome, Xfce, etc. etc. desktop, apps, etc. in all "classical" BSDs because the software stack they share is a lot more than the software stack shared with Mac OS X. Actually, though the "classical BSDs" and Mac OS X have the same heritage, I think the BSDs and Linux are closer each other.

So, do not know... the human is, like the monkeys, a primate; but based on that, you cannot state that the human is a monkey.

Edited 2011-08-11 04:19 UTC

Reply Parent Score: 2

RE: Ready to Rumble?
by _xmv on Thu 11th Aug 2011 00:13 in reply to "Ready to Rumble?"
_xmv Member since:
2008-12-09

In the left corner: We have the Linux User, denying that OSX is still a BSD.

In the right corner: The BSD Users, claiming that with OSX makes BSD more popular.

Lets get ready to rumble ;)

No, Serious.
I think OSX is definite part of the BSD Family. I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.

Linux is a kernel.
BSD is an operating system.
GNU/Linux is an operating system. Android is certainly not a GNU/Linux. It's just some weird stuff running on the Linux kernel.

Reply Parent Score: 4

RE: Ready to Rumble?
by Soulbender on Thu 11th Aug 2011 14:53 in reply to "Ready to Rumble?"
Soulbender Member since:
2005-08-18

I mean, when you call "android" a linux, then OSX is a BSD.


It runs on a Linux kernel so obviously it is a Linux.

Reply Parent Score: 3