Re: SOS!!Boot Loader StartUp Function !!

baddg wrote:
I know that bootloader can be run as different stage.For instance,stage
1 to init the cpu (asm code)and stage 2 to do sth else(c code).I was
told recently that at the end of stage1 run in flash ,stage2 code was
copy to RAM and JUMP to that address .So Bootloader stage 2 can be run
in RAM.I wonder How WinCE Bootloader deal with it?
If you want to jump to RAM you had better copy the RAMIMAGE section into RAM first, it will not magically move itself out of flash.


question2:In Eboot.bib file,

; Name Start Size Type
; ------- -------- -------- ----
STACK 80000000 00010000 RESERVED
RAM 80010000 00010000 RAM
EBOOT 80020000 00040000 RAMIMAGE
RSVD 80060000 0009F000 RESERVED
ARGS 80FF0000 00001000 RESERVED
the flash starts at 0x0000,0000 and the size of it is 32M
the RAM starts at 0xA000,0000 and the size of it is 64M
In g_oalAddressTable,
DCD 0x80000000, 0xA0000000, 64 why the size of RAMIMAGE is 256K?what does this mean?
help me.

Read about it in the help files here:
Developing an Operating System > Build System > OS Design Configuration Files > Run-Time Image Configuration Files > Binary Image Builder File

Look at the MEMORY SECTION information.

256K is used as RAMIMAGE space, this is where the code and initialized static data live for your eboot image. It is 256K because eboot probably fits into 256K.

Geoff
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