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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RE: Ready to Rumble?
by Vanders on Wed 10th Aug 2011 21:07 UTC 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