Huawei Factory Unlock Code V5 Best
: Place a SIM card from a "prohibited" network into the device.
: Factory codes sourced from servers can take anywhere from a few hours to 10 business days to deliver. 🛠️ How to Enter the Code huawei factory unlock code v5
Huawei discontinued its official bootloader unlocking service on July 25, 2018 : Place a SIM card from a "prohibited"
The Huawei Factory Unlock Code V5 has significant implications for device security and user freedom: huawei factory unlock code v5
Nice write up – where can I get the vulnerable app? I checked IOLO’s website and the exploitdb but I can’t find 5.0.0.136
For “System Shield AntiVirus and AntiSpyware” you’ll need to run the downloader which downloads the main installation package but then you’ll need to also request a license. Best just to download “System Mechanic Pro” and install as a trial, this downloads the entire package and no license is required for installation
http://download.iolo.net/sm/15/pro/en/iolo/trial/SystemMechanicPro_15.5.0.61.exe
Hello.
Thanks for this demonstration!
I have a question. With this exploit, can we access to the winlogon.exe and open a handle for read and write memory?
Kind regards,
Yes you can as “SeDebugPrivilege” is also enabled
Why doesn’t it work with csrss.exe?
pHandle = OpenProcess(PROCESS_VM_READ, 0, 428); //my csrss PID
printf(“> pHandle: %d || %s\n”, pHandle, pHandle);
i got: 0 || (null)
It should work, most likely haven’t got the necessary privilege
Oh yes, thanks. But can you help me with “SeDebugPrivilege”. What offset?
Kind regards,
The SeDebugPrivilege is already enabled in this exploit, what you can do it use a previous exploit of mine which uses shellcode being injected in the winlogon process.
Thanks for nice write up. I want to study this case, so I’ve downloaded the link
http://download.iolo.net/sm/15/pro/en/iolo/trial/SystemMechanicPro_15.5.0.61.exe.
And opened amp.sys file with IDA pro, but I could not find the code related to ctl code 0x00226003. How can I find it?
Best just do a text search for 226003 and only one entry will be listed
Thanks! I found with its hex byte ’03 60 22′ in IDA search and reached vulnerable function.