Wolfgang Denk wrote:
Sorry, I have no idea. If really needed, I can arrange to (temporarily) put a copy of the U-Boot archive on our FTP server (the bzip2ed tarball for U-Boot-Users is 22 MB, and 40 MB for PPCBoot).
OK, well I may take you up on that but I'll get back to you. I hate having to ask questions I know have been answered already.
So try:
bash$ mkimage -T multi -C gzip -a 0 -e 0 -n 'test multifile' \ -d /path/to/kernel_image:/path/to/ramdisk_image uMulti
Hmmm... include/image.h talks about having byte counts in there as well. Maybe I'm just confused. I'll recheck it.
See the README (section "More About U-Boot Image Types") for the file format. This explains that with a mutlifile image with a kernel and a ramdisk you will have to skip a total of 76 bytes of header information (64 for U-Boot header + 4 for kernel size + 4 for ramdisk size + 4 for terminating null).
Thank you. Very helpful. Can't believe I missed the README reference! Ugh!
So I recommend a three step approach:
bash$ mkimage -l uMulti Image Name: Linux-2.4.4-2003-04-05 Multiboot Created: Sun Apr 6 12:44:18 2003 Image Type: PowerPC Linux Multi-File Image (gzip compressed) Data Size: 2610664 Bytes = 2549.48 kB = 2.49 MB Load Address: 0x00000000 Entry Point: 0x00000000 Contents: Image 0: 715859 Bytes = 699 kB = 0 MB Image 1: 1894792 Bytes = 1850 kB = 1 MB
bash$ dd if=uMulti bs=76 skip=1 of=foo 34350+1 records in 34350+1 records out bash$ dd if=foo of=kernel.gz bs=715859 count=1 1+0 records in 1+0 records out bash$ gzip -vt kernel.gz kernel.gz: OK
The "bs=715859" uses the kernel data size as printed by the "mkimage -l" command.
Very much appreciated. Thanks again!
Tom