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Re: Script to create bootable arm images?



> On Jun 24, 2022, at 12:45 PM, Jason Thorpe <thorpej%me.com@localhost> wrote:
> 
>> On Jun 24, 2022, at 11:02 AM, Brook Milligan <brook%nmsu.edu@localhost> wrote:
>> 
>> build.sh works great to create, for example, binary/gzimg/armv7.img.gz.  
>> 
>> However, that is not necessarily a bootable image, at least on some systems.  In addition, various u-boot magic files must be added to the FAT partition at the beginning of armv7.img to make it bootable.
> 
> installboot(8) (and the tool-ified version) has support for plopping the correct uboot into the image.  You need to install the uboot packages first.

Thanks.  So all that is needed is the following (for cross compiling):

gzip -c -d < /path/to/release/evbarm-earmv7hf/binary/gzimg/armv7.img.gz > /tmp/armv7.img
env INSTALLBOOT_UBOOT_PATHS=/path/to/pkg/share/u-boot  /path/to/tools/bin/nbinstallboot  -m evbarm -o board=ti,am335x-bone-black  /tmp/armv7.img

Does it make sense to consider a command line option for installboot that sets the u-boot path so the environment variable is not needed?

In answer to my original question, I think this means that all the images on armbsd.org can be created with a loop over the various boards with each iteration running the two commands above.  Is that correct?  Is that what is happening?

Thanks for your help.

Cheers,
Brook






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