Jailbreak Source Code Revealed for iOS 8.4.1 iPhone
The long awaited iOS 8.4.1 jailbreak release date might be this year thanks to Luca Todesco. This developer decided to unveil the source code for iOS 8.4.1 jailbreak. This is the last version of iOS 8 released by Apple. The company offers iOS 9, 9.0.1 and 9.0.2 now.
Hackers were not able to find new exploits in iOS 8.4.1 and newer mobile versions for iPhone, iPad and iPod touch. The Cupertino-based company managed to kill all the exploits used in iOS 8.4 and earlier firmwares that can be jailbroken.
At the same time, hacker iH8sn0w demoed iOS 9 jailbreak on his iDevice however he has no intensions to make the tool public. iOS 8.4.1 jailbreak sources might help other groups of hackers to find the exploits and create a new program that will allow public jailbreaking their smartphones and tablets.
It could be important to downgrade from iOS 9 to 8.4.1 while Apple is supporting this firmware version. This way you’ll be able to jailbreak once such an option becomes available. Experts believe that the code can help the release to happen.
The source code jailbreak iPhone that Todesco shared with users is described as the ‘incomplete iOS 8.4.1 jailbreak’ tool. There is one big problem with source codes. They are mostly limited to tethered releases and they are not as popular and untethered iOS 8.4.1 jailbreak. It is required to run the program each time the gadget reboots and this can lead to user’s headache.
Most hackers are trying to develop an untethered program that can automatically perform the jailbreak and stays on the iPhone until new firmware upgrade.
Developers can surely use the source code for performing this procedure however ordinary users will not be able to jailbreak properly. Besides, incomplete jailbreaking doesn’t result in Cydia installation and Cydia store is number one reason why iPhone and iPad owners choose to jailbreak. This is the place with tons of jailbreak tweaks and customization options that make the iDevice experience so much better.