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Just as the question/title says. I understand that running Mac OS inside of a VM is against the EULA for the consumer version (but not the server, which is much more expensive!) If I were to purchase a legal copy of Mac OS, and install it to a VM, then register as an Apple Developer, would they shut me out? Is there a way they can tell the difference between emulated hardware and Apple computers?

I'm slightly unfamiliar with how all of Apple's software works. Windows goes through this "genuine" test whenever installing service packs, but I don't know if Mac goes through the same trouble.

Many thanks,

-Tom

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2 Answers 2

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To the best of my knowledge, Mac OS X does not "phone home" to report hardware and it certainly doesn't have any activation schemes of any kind.

But hey, Apple employees might be on superuser. ;-)

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I have no plans to purchase anything Apple, I am asking this in this place of a community which is apprehensive about the cost of developing on Apple products. Thanks for the reply, nonetheless, and merry Christmas. :) –  Thomas Havlik Dec 22 '10 at 6:10
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If you have no plans to purchase anything from Apple, you won't be breaking a EULA, you'll be pirating software. So why bother asking the question? :) –  ghoppe Dec 22 '10 at 6:11
    
Mac OS X does absolutely nothing to verify it's installed on a "real" Mac. They also only charge update price for Mac OS X 10.6 although it's the full version and can, theoretically of course, be installed on a pristine hard disk. –  Daniel Beck Dec 22 '10 at 6:15
    
No, I won't be pirating it. I won't be buying it, or using it. –  Thomas Havlik Dec 22 '10 at 6:51

Mac OS X doesn't have any kind of serial numbers or other anti-piracy protections, but it will not run virtualized by default, you'd likely need to install a hackintosh version. I'm not fully aware if any part of the iOS developer signing stack is tied to the hardware, but I doubt it.

And yes, an OS can trivially detect if it's running inside a virtual machine monitor, usually using some form of a timing attack.

EDIT: I'm not sure if OS X does, however, employ such a method of detection.

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