RE: [Ltib] generate uImage instead of zImage

[Stuart Hughes]

Hi Kevin,

On Mon, 2009-01-12 at 22:24 +0100, Kevin Wells wrote:
> I'm looking at this also. Using u-boot, I need a uImage type.
> 
> I've added a SYSCFG_KTARG with a default "uImage" value and I do see the make 
> command pass "uImage" to the kernel build.
> However, with the ARM arch, I'm still not getting a uImage type.
> 

You need to check your kernel sources (and the mailing lists) to make
sure that for arm you have Makefiles that will build a uImage.

> Similar to the several m68k platforms that require special uImage processing, 
> I've added a line similar to the following in the common kernel spec 
> template..
> 
> if [ $CPU = LPC3XXX ]
> then
>     mkimage -A arm -O linux -T kernel -C none -a 0x80008000 -e 0x80008000 -n 
> "Linux kernel Image" -D$RPM_BUILD_ROOT/%(pfx)/boot/zImage 
> $rpm_build_root/%(pfx)/boot/uImage
> fi
> 

These #ifdef $CPU clauses are deprecated, I've mentioned to the m68k
guys before that they should not need to do this.  Please avoid this if
you can (remove when you can).

> Is there a way to get the uImage built for the ARM target with those 
> addresses without modifying the common template?

I think the solution is to fix the kernel sources so they properly deal
with the make target uImage.  This is a question for the kernel mailing
list.

Regards, Stuart