Hex File Format Question

[eepty]

Hi I am working on a bootloader and an application code for a microcontroller. I have finished both and they generated two seperated hex files. I know I can cascade the application hex after the bootloader hex so that I have a single hex file and burn it into the microcontroller.

The ending part of the hex on the bootloader:

:10287C00000000000000000000000000000000004C
:10288C00000000000000000000000000000000003C
:10289C00000000000000000000000000000000002C
:0400000300001CA934
:00000001FF

And on the application hex:

:10457C00A8080020A8080020B0080020B0080020DF
:10458C00B8080020B8080020C0080020C00800208F
:10459C00C8080020C8080020FFFFFFFF0000020031
:040000031000717DFB
:00000001FF

When I cascade the two hex files, on the bootloader hex, I have tried to
1.)just remove the last line:

:00000001FF

2.) or remove the last two lines:

:0400000300001CA934
:00000001FF

I found both method works...I know the meanning of

:00000001FF

is just "end of file" so I have to remove it. But what is the meaning of the second-to-last line?

:040000031000717DFB

 

[Gorgon]

It's the 'start segment address' record(type 03).
How this is related to your microprocessor I don't know. I would assume it defines the starting point of your program.

This 'Intel hex' format relates in original to the 80x86, and states the CS:IP pair of 4 bytes.