How to define UEFI entry point when image is loaded in RAM by ARM trusted FW.

Hello

I would  like to have your help on Uefi boot on ArmV8 platform (aarch64).

I’m trying to setup the entire boot on my platform based on Armv8 processors with Arm Trusted Firmware and UEFI (from linaro-edk2).

I’m already running the Arm Trusted Firmware successfully, and now I try to run the Uefi-edk2 from Linaro.

I created a new platform/package in ArmPlatformPkg mainly inspired from ArmJunoPkg and ArmVEpessPkg/ArmVExpress-FVP-AArch64.

I succeeded to generate a uefi binary image for my platform but I have an issue.

In the boot process, the ARM Trusted FW copies the entire UEFI image in RAM and SW jumps at the first address of the image.

On Juno and FVP foundation, I see that at the first address of the image, there is a jump instruction to the entry point of the UEFI.

The problem is that on my image, there are zero instead of the jump instruction, then I cannot start properly to run the UEFI.

I guess this is GenFds that patches the binary to add the jump instruction, but I do not understand where is my error in configuration files or what is missing to have this jump instruction. Where is the entry point defined ?

Any help or pointers to information are welcome.

Thanks in advance

Regards

Nicolas