On Fri, Apr 25, 2008 at 02:31:20PM +0530, sahlot arvind 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???
It's all rather mystical if you don't understand the ARM instruction set.
1. 'b' is a branch instruction. All branches are relative to the
current PC.
2. the code is not executed in the location where you find it in
the kernel. It is copied to other locations in memory. Other
code (between __stubs_start and __stubs_end) is copied to 512
bytes above the start of the vectors, and these branch instructions
branch to that other code.
3. __stubs_offset is just a correction factor to convert the branches
to point at the correct _relative_ position when they're copied to
their proper location in memory.
The simple way to _read_ the code is to ignore the 'stubs_offset' and just
follow the branch instruction to vector_irq:
/*
* Interrupt dispatcher
*/
vector_stub irq, IRQ_MODE, 4
and then look at what vector_stub expands to.
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