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Recently, iOS 6 was "jailbroken" but only on the Apple A4 CPU.

Why is the "jailbreaking" process specific to a CPU?

From Wikipedia:

... "iOS jailbreaking is the process of removing the limitations imposed by Apple on devices running the iOS operating system through the use of hardware/software exploits – such devices include the iPhone, iPod touch, iPad, and second generation Apple TV. Jailbreaking allows iOS users to gain root access to the operating system""" ...

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apple does things in the name of DRM that shouldn't be done in good software development –  Ryathal Sep 20 '12 at 15:18
    
The security model starts with the CPU. It will only boot into code signed by Apple so no one else can put their own bootloader on the device. –  Guy Sirton Sep 20 '12 at 17:57
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@Ryathal: Isn't that a redundancy? –  Mason Wheeler Sep 20 '12 at 18:23
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1 Answer

Odds are that the older A4 chips have an instruction which may or may not be undocumented that allows for an exploit that enables a rooting to be performed. It is possible that that exploit became known during the design process for the next generation of ARM chips for Apple devices and was either removed or otherwise disabled.

Also the longer a given device has been on the public market, the more time that the various cracking groups have had to take a look at and dissect the hardware for that device. Once you map the circuitry for a chip fully, understanding what it does is substantially easier. This process can take a long time in many cases given the size (smallness) and complexity (staggering) of modern integrated circuits.

Much of the analysis involved (and the design of the devices) are machine driven at this point to deal with the complexities inherent. Thus a new chip may have a totally different architecture requiring an entirely different approach to rooting or otherwise exploiting it.

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