Dear Tom,

in message 3F4112FD.6080204@san.rr.com you wrote:

now on the SoureForge page! Is there any place I can dowload the archives?

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).

I haven't been successful referencing the include/image.h file on how to use "mkimage -T multi". Any examples out there? Basically I just want to create an U-Boot image of a gzipped kernel and ramdisk.

"mkimage" will print a help message when called without arguments:

bash$ mkimage Usage: mkimage -l image -l ==> list image header information mkimage -A arch -O os -T type -C comp -a addr -e ep -n name -d data_file[:data_file...] image -A ==> set architecture to 'arch' -O ==> set operating system to 'os' -T ==> set image type to 'type' -C ==> set compression type 'comp' -a ==> set load address to 'addr' (hex) -e ==> set entry point to 'ep' (hex) -n ==> set image name to 'name' -d ==> use image data from 'datafile'

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

Also, I received some pointers on how to extract the U-Boot header from an image (thanks Wolfgang and Marc Singer!). Now I'm going to want to see if I can extract the kernel from a multi-file image. I think this will be similar to the header extract but any pointers are greatly appreciated. Fightin the clock on this one!

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).

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.

Hope this helps.

Best regards,

Wolfgang Denk