0

I am trying to build a LineageOS ROM for an unsupported device, i need to get the partition size of the device, I have a rooted device and then went to /proc/partitions via root explorer, and I have like-

    major minor #blocks name
  • 7 0 65536 loop0
  • 7 1 98304 loop1
  • 179 0 122138624 mmcblk0
  • 179 1 16384 mmcblk0p1
  • 179 2 16384 mmcblk0p2
  • 179 3 16384 mmcblk0p3
  • 179 4 4096 mmcblk0p4
  • 179 5 16384 mmcblk0p5
  • 179 6 16384 mmcblk0p6
  • 179 7 2048 mmcblk0p7
  • 179 8 1024 mmcblk0p8
  • 179 9 1024 mmcblk0p9
  • 179 10 40960 mmcblk0p10
  • 179 11 8192 mmcblk0p11
  • 179 12 32768 mmcblk0p12
  • 179 13 8192 mmcblk0p13
  • 179 14 65536 mmcblk0p14
  • 179 15 262144 mmcblk0p15
  • 259 0 307200 mmcblk0p16
  • 259 1 32768 mmcblk0p17
  • 259 2 2359296 mmcblk0p18
  • 259 3 118931419 mmcblk0p19
  • 179 48 4096 mmcblk0rpmb
  • 179 32 4096 mmcblk0boot1
  • 179 16 4096 mmcblk0boot0
  • 179 64 7761920 mmcblk1li
  • 179 65 7757824 mmcblk1p1

I want to under the in which format these block sizes are. whether they are in bytes, kb or mb?

As I know mmcblk0p1 is the boot partition and its block size is 16384, so what it will be in bytes?

  • 1
    Yeah the sizes are in bytes – esQmo_ May 27 '18 at 12:20
  • @esQmo_ Do proper math? 16384B is just 16KB. // To OP: from my (admittedly shallow) Linux knowledge, reported block size is always 512 bytes. That gives 8MB for mmcblk0p1, and ~58GB for mmcblk0 (the entire eMMC). Is your device 64GB? If yes this would be the reasonable amount. – Andy Yan May 27 '18 at 12:37
  • I have doubt as 1 MB is almost equals to 1048576, then it means here boot partition is of 0.016384 MB – Rahul Sharma May 27 '18 at 12:41
  • @AndyYan i have an 128gb device – Rahul Sharma May 27 '18 at 13:20
  • @Andy Yan The size is indeed 16 mb as 16384kb. Should've I just said approximately 17mb :p – esQmo_ May 27 '18 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.