Running app on iOS 9.3.3 jailbroken with root access

Question

I am trying to give root access to my app on Jailbroken iPhone 9.3.3

I have followed the same steps as mentioned here - Gaining root permissions on iOS for NSFileManager (Jailbreak)

However the app does not gets root access. It looks like, the setuid(0) fails. I am using setuid(0) and setgid(0) in main function.

int main(int argc, char *argv[]) {
   if(!(setuid(0) == 0 && setgid(0) == 0)) {
     exit(EXIT_FAILURE);
   }

   return UIApplicationMain(argc, argv, nil, NSStringFromClass([AppDelegate class]));
}

The app exits in the above condition.

Any help will be greatly appreciated.

On your device, could you please run `stat -c '%U %a' path/to/wrapper-script /path/to/executable` and post the output here? — Siguza, Aug 02 '16 at 18:15
stat: command not found. Do I have to look for this command in particular directory? I am hitting it from root. — Michael, Aug 02 '16 at 19:45
My bad, you're gonna need the "Core Utilities" package from Cydia. I thought that was installed by default. — Siguza, Aug 02 '16 at 19:51
for the script it gives `root 775` and for the executable it gives `root 6775`. — Michael, Aug 02 '16 at 20:39

score 0 · Answer 1 · answered Sep 05 '16 at 01:38

setgid(0) is not necessary

In iOS you can't run apps as root directly you have to run them through a shell script.

Todo that:

Rename the app binary, for example if the binary named App rename it to _App
Create a text file named App "the name of the original binary name"
Inside the new created text file paste

#!/bin/bash
dir=$(dirname "$0")
exec "${dir}"/_App "$@"

Running app on iOS 9.3.3 jailbroken with root access

1 Answers1