On Mon, Dec 12, 2011 at 03:31:30AM +0100, Andi Kleen wrote:

quoted

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@fromorbit.com

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