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
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
and the SAVE_ARGS_IRQ macro switches to the per cpu interrupt stack.
The only caller does this:
addq $-0x80,(%rsp) /* Adjust vector to [-256,-1] range */
So, why do we get this:
Dec 6 20:27:55 localhost kernel: <IRQ> [<ffffffff81067097>] ?
Dec 6 20:27:55 localhost kernel: [<ffffffff8106f6da>] ?
Dec 6 20:27:55 localhost kernel: [<ffffffff81067186>] ?
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>] ?
at the top of the stack frame? Is the stack unwinder walking back
across the interrupt stack to the previous task stack?