bash hex 14digits incremental (yes, it's a bruteforce)

Question

I am trying to make a bash cript to bruteforce attack an android oem bootloader key to unlock the fastboot. I made this script

#!/usr/bin/zsh
setopt c_bases
typeset -i 16 num
for num in $( seq $((0x10000000000000)) 1 $((0xffffffffffffff)) ); do 
    echo "$num"
    fastboot oem unlock $num
done

but i have problems because iif i try to execute it i run out of ram. Is there any way to make an incremental script like this without running out of memory? Thanks a lot

On my 24 core Haswell system, I can fork ~20k programs per second. If `fastboot` doesn't need any time to talk to the device and is perfectly parallelizable, this would take me 100,000 years. Good luck! — that other guy, Apr 03 '17 at 17:49
Well, i dn't have so much spare tine, so i will have to ask for the unlocking code... Thanks antway! — Filippo Falezza, Apr 05 '17 at 05:48

score 2 · Answer 1 · answered Apr 03 '17 at 16:59

for num in $( seq $((0x10000000000000)) 1 $((0xffffffffffffff)) ); do

will never work, because the entire command substitution output will be placed in memory before the loop is invoked. Use a c-style loop (bash specific) for loops with many iterations:

for ((num=0x10000000000000;num<=0xffffffffffffff;num++)); do

You may also consider using gnu parallel to improve the speed of the test.

bash hex 14digits incremental (yes, it's a bruteforce)

1 Answers1