lpmake requires to get the size of the “super” partition on the device:

-d,--device-size - The size of the “super” partition on the device. It must match exactly, and it must be evenly divisible by the sector size (512 bytes).

How can I obtain it?

Using adb, I was able to run ls -la /dev/block/bootdevice/by-name and see

lrwxrwxrwx 1 root root 21 2022-06-13 23:32 super -> /dev/block/mmcblk0p34

and then using termux, as root, I obtained:

blockdev --getsz /dev/block/mmcblk0p34

but since the original firmware's system.img file is 3.2 GiB (3,435,159,228 > 8,388,608), this seems quite unlikely.

What am I missing? I am aware of this post, but it did not helped.

The scatter file reads

- partition_index: SYS35
  partition_name: super
  file_name: super.img
  is_download: true
  type: NORMAL_ROM
  linear_start_addr: 0x23800000
  physical_start_addr: 0x23800000
  partition_size: 0x100000000
  region: EMMC_USER
  storage: HW_STORAGE_EMMC
  boundary_check: true
  is_reserved: false
  operation_type: UPDATE
  is_upgradable: true
  empty_boot_needed: false
  reserve: 0x00

and 0x10000000 is 268435456 (268,435,456), but that's still not enough?

I see another number using fastboot...

fastboot getvar partition-size:super
partition-size:super: 100000000
finished. total time: 0.003s


echo "ibase=16; 100000000"|bc

This one could be valid, since 4,294,967,296 > 3,435,159,228.

  • 1
    blockdev --getsize64 8388608 (blocks) x 512 (block size) = 4294967296 = 0x100000000 (not sure where you got 0x10000000 its not mentioned in scatter file)
    – alecxs
    Commented Jun 15, 2022 at 18:48
  • Thanks a lot, I had forgotten a 0 in my conversion…
    – Clément
    Commented Jun 15, 2022 at 21:17

1 Answer 1


Actually, all three sources are in agreement when you know how to read it:


partition_size: 0x100000000

0x100000000 = 4,294,967,296 (I had forgotten a 0 in my question)


blockdev --getsz /dev/block/mmcblk0p34

8388608 = 8388608 (blocks) x 512 (block size) = 4,294,967,296 (Thanks alecxs!)


fastboot getvar partition-size:super

Gives 0x100000000 = 0x100000000 = 4,294,967,296

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