Subject: entry_point

From: [email-masked]
Date: Tue, 3 Aug 1999 11:32:36 +0100

Hello,

     Here is some 'basic questions":
This is what my "System.map" looks like. (only the beginning) ( I have a
brutus-like board, and I load the kernel into the RAM)

00000000 A _text
c0004000 A swapper_pg_dir
c0008000 A __init_begin
c0008000 T _stext
c0008000 T stext
c000802c t __entry

Does that mean the kernel (vmlinux) must be loaded at c0008000, and carried out
at c000802c ?
( I mean, from the board, when I want to carry out the kernel, I must have
something like "call c000802c"?)
Is it possible to have the loading address and the executing address both at
c0000000 ?

Thanks
Arnaud

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From: Philip Blundell <[email-masked]>
Date: Tue, 03 Aug 1999 11:56:46 +0100

>Does that mean the kernel (vmlinux) must be loaded at c0008000, and carried 
>out at c000802c ?

You should indeed load at 0xc0008000.  On the SA-1100 you end up with the 
Angel bootstrap code at 0xc0008000 and the real kernel entry at 0xc000802c.  I 
think the intention is that you always start executing at 0xc0008000; if you 
don't use Angel you probably ought to fiddle the code so that the appropriate 
bits are left out.

In a flat kernel binary (Image/zImage) there is no symbol or entry-address 
information so starting at the first instruction is pretty much the only 
option.

>Is it possible to have the loading address and the executing address both at

It's possible but it wouldn't be completely trivial.  Why do you want to do 
this?

p.