MIMXRT1040 - L2Boot launch App

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MIMXRT1040 - L2Boot launch App

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riccardo_carlot
Contributor III

Hello,

I'm facing with my custom bootloader on XRT1040 device.
This L2 bootloader, at one point, will have to choose the right application image to execute and launch it. I supposed to have two images saved in the flash memory.

The situation could be something like this:

riccardo_carlot_0-1701267469718.pngriccardo_carlot_0-1701267469718.png

For istance, I located APP_A at 0x00001000 and APP_B at 0x00002000.

My dubt is about the bootloader, when it jumps to one of the two images.
Since both aplication are compiled and builted as stanalone applications, starting from address 0x00000000, how can I add a costant offset to the application that executes after the boot?.
For example, both APP_A and APP_B are builted starting from 0x00000000, how can I say to the device that it have to add 0x00001000 when it executes the APP_A or 0x00002000 when it executes the APP_B?

I know I have to move the VTOR, but this register is about the reset vector, it doesn't touch the fetching process. 

Thank in advance.

Riccardo

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jingpan
NXP TechSupport
NXP TechSupport

Hi @riccardo_carlot ,

RT1040 has three registers IOMUXC_GPR_GPR30, IOMUXC_GPR_GPR31 and IOMUXC_GPR_GPR32 which are used to remap part of flexspi address.

 

Regards,

Jing

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jingpan
NXP TechSupport
NXP TechSupport

Hi @riccardo_carlot ,

RT1040 has three registers IOMUXC_GPR_GPR30, IOMUXC_GPR_GPR31 and IOMUXC_GPR_GPR32 which are used to remap part of flexspi address.

 

Regards,

Jing

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