On Mon, Dec 12, 2011 at 03:31:30AM +0100, Andi Kleen wrote:
> > But that happens before do_IRQ is called, so what is the do_IRQ call
> > chain doing on this stack given that we've already supposed to have
> > switched to the interrupt stack before do_IRQ is called?
>
> Not sure I understand the question.
>
> The pt_regs are on the original stack (but they are quite small), all the rest
It's ~180 bytes, so it's not really that small.
> is on the new stack. ISTs are not used for interrupts, only for
> some special exceptions.
IST = ???
> do_IRQ doesn't switch any stacks on 64bit.
No, but it appears that it's caller does:
/* 0(%rsp): ~(interrupt number) */
.macro interrupt func
/* reserve pt_regs for scratch regs and rbp */
subq $ORIG_RAX-RBP, %rsp
CFI_ADJUST_CFA_OFFSET ORIG_RAX-RBP
SAVE_ARGS_IRQ
call \func
.endm
and the SAVE_ARGS_IRQ macro switches to the per cpu interrupt stack.
The only caller does this:
common_interrupt:
XCPT_FRAME
addq $-0x80,(%rsp) /* Adjust vector to [-256,-1] range */
interrupt do_IRQ
So, why do we get this:
Dec 6 20:27:55 localhost kernel: <IRQ> [<ffffffff81067097>] ? warn_slowpath_common+0x87/0xc0
Dec 6 20:27:55 localhost kernel: [<ffffffff8106f6da>] ? __do_softirq+0x11a/0x1d0
Dec 6 20:27:55 localhost kernel: [<ffffffff81067186>] ? warn_slowpath_fmt+0x46/0x50
Dec 6 20:27:55 localhost kernel: [<ffffffff8100c2cc>] ? call_softirq+0x1c/0x30
Dec 6 20:27:55 localhost kernel: [<ffffffff8100dfcf>] ? handle_irq+0x8f/0xa0
Dec 6 20:27:55 localhost kernel: [<ffffffff814e310c>] ? do_IRQ+0x6c/0xf0
Dec 6 20:27:55 localhost kernel: [<ffffffff8100bad3>] ? ret_from_intr+0x0/0x11
Dec 6 20:27:55 localhost kernel: <EOI> [<ffffffff8115b80f>] ? kmem_cache_free+0xbf/0x2b0
at the top of the stack frame? Is the stack unwinder walking back
across the interrupt stack to the previous task stack?
Cheers,
Dave.
--
Dave Chinner
david@xxxxxxxxxxxxx
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