Re: Question on flow of control at the time of interrupt.

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On Fri, Apr 25, 2008 at 5:01 PM, sahlot arvind <asahlot@xxxxxxxxx> wrote:
> Guys,
>
>  I am trying to understand the flow of control when an interrupt comes.
>  I am reading linux-2.6.24 src code and looking at
>  arch/arm/kernel/entry-armv.S.
>  At the bottom of this file I see the vector table as below -
>
>  __vectors_start:
>     swi SYS_ERROR0
>     b   vector_und + stubs_offset
>     ldr pc, .LCvswi + stubs_offset
>     b   vector_pabt + stubs_offset
>     b   vector_dabt + stubs_offset
>     b   vector_addrexcptn + stubs_offset
>     b   vector_irq + stubs_offset
>     b   vector_fiq + stubs_offset
>
>     .globl  __vectors_end
>
>
>  Here is not 'stubs_offset' a constant? and after seeing an IRQ where are we
>  branching by doing ' b   vector_irq + stubs_offset'  and what is the flow of
>  control???
>  Thanks in advance.
>

stubs_offset is a constant decided by the assembler, note the .equ
above. So actually, the vector_irq + stubs_offset (for example)

= vector_irq + __vectors_start + 0x200 - __stubs_start
= (vector_irq - __stubs_start) + 0x200 + __vectors_start

And at run-time there will be dedicated page allocated for the
vectors and the stubs code, and the stubs code will be copied
at a offset of 0x200.

see? 1st part is the vector_irq routine offset relative to the start of
__stubs_start, __vectors_start is used to compensate for the
assembler to generate relative branch instruction.


>  - A Sahlot
>  --
>  http://linuxexplained.blogspot.com
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-- 
Cheers
- eric

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