Hello friends,
I read some materials from internet.
"Extract an uImage
According to the U-Boot header definition an uImage begins with the hex byte sequence 27 05 19 56 and the header is 64 bytes long.
The data is directly following the header and could be in compressed form (gzip or bzip2), but this is untypical for embedded devices. It is normally a zImage (=compressed Linux kernel plus decompressor for low memory). To get the data only the uImage header has to be removed.
For ARM devices an additional 8-byte header with the machine type is in front of a zImage. In this case the additional 8 bytes have to be removed too."
but I don't know where is the 8-byte header with the machine type?
here is my case:
Code:
root@benjamin-virtual-machine:/GPL/tmp/uimage# mkimage -l uImage\ v5.2
Image Name: Linux-2.6.32.9
Created: Tue Sep 27 16:47:13 2011
Image Type: ARM Linux Kernel Image (uncompressed)
Data Size: 2448712 Bytes = 2391.32 kB = 2.34 MB
Load Address: 40008000
Entry Point: 40008000
root@benjamin-virtual-machine:/GPL/tmp/uimage# hexdump -C uImage\ v5.2 | more
00000000 27 05 19 56 03 ec 7f 58 4e 81 8d 91 00 25 5d 48 |'..V...XN....%]H|
00000010 40 00 80 00 40 00 80 00 30 cb 44 69 05 02 02 00 |@...@...0.Di....|
00000020 4c 69 6e 75 78 2d 32 2e 36 2e 33 32 2e 39 00 00 |Linux-2.6.32.9..|
00000030 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
00000040 00 00 a0 e1 00 00 a0 e1 00 00 a0 e1 00 00 a0 e1 |................|
*
00000060 02 00 00 ea 18 28 6f 01 00 00 00 00 48 5d 25 00 |.....(o.....H]%.|
00000070 01 70 a0 e1 02 80 a0 e1 00 20 0f e1 03 00 12 e3 |.p....... ......|
00000080 01 00 00 1a 17 00 a0 e3 56 34 12 ef 00 20 0f e1 |........V4... ..|
00000090 c0 20 82 e3 02 f0 21 e1 00 00 00 00 00 00 00 00 |. ....!.........|
000000a0 d4 00 8f e2 7e 30 90 e8 01 00 50 e0 0a 00 00 0a |....~0....P.....|
Any help will be appreciated.
