>/*
> * Interrupt dispatcher
>*/
> vector_stub irq, IRQ_MODE, 4
After expansion it will look some thing like this -
vector_irq:
...
blah - blah
...
@ the branch table must immediately follow this code
and lr, lr, #0x0f
mov r0, sp
ldr lr, [pc, lr, lsl #2]
movs pc, lr @ branch to handler in SVC mode
.long __irq_invalid @ 1 (FIQ_26 / FIQ_32)
. .long __irq_invalid @ 2 (IRQ_26 / IRQ_32)
.long __irq_svc @ 3 (SVC_26 / SVC_32)
.long __irq_invalid @ 4
.long __irq_invalid @ 5
.long __irq_invalid @ 6
.long __irq_invalid @ 7
.long __irq_invalid @ 8
.long __irq_invalid @ 9
.long __irq_invalid @ a
.long __irq_invalid @ b
.long __irq_invalid @ c
.long __irq_invalid @ d
.long __irq_invalid @ e
.long __irq_invalid @ f
I guess that the instruction in the above code -
movs pc, lr @ branch to handler in SVC mode
will jump to address specified by
.long __irq_svc @ 3 (SVC_26 / SVC_32)
Assuming I am right then the instruction -
ldr lr, [pc, lr, lsl #2]will set 'lr' to the address of '__Irq_svc'. But I am unable to understand how come this instruction is setting up lr to point to __irq_svc?
In the above code while executing 'and' instruction 'lr' contains the value of 'spsr' and last four bits are mode bits then this 'and' instruction is preserving those 4 bits (mode bits). But why do we need those bits and in 'ldr' why we are doing left shift by 2? Cannot we simply add 16 (since there are one instruction and 3 handler addresses after 'ldr' instruction) to 'pc' and load it into 'lr' so that 'lr' starts pointing to __irq_svc, which is 4 words down the 'ldr' instruction?
And one more question why do we have 16 handlers in each exception stub? why not just 1 (e.g. __irq_svc) or 2 (__irq_usr and __irq_svc)? and what is use of __irq_usr?
I am sorry I have put lot of questions in this email. But answers to them would really help me in understanding the code.
Thanks in advance.
- A Sahlot
On Sat, Apr 26, 2008 at 11:37 AM, sahlot arvind <asahlot@xxxxxxxxx> wrote:
Thanks Russell. I will re-try and get back in case dont understand.- A--On Sat, Apr 26, 2008 at 2:09 AM, Russell King - ARM Linux <linux@xxxxxxxxxxxxxxxx> wrote:
It's all rather mystical if you don't understand the ARM instruction set.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???
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.
http://linuxexplained.blogspot.com
--
http://linuxexplained.blogspot.com
