Messages in this thread | | | Date | Tue, 7 Aug 2012 17:15:31 +0200 | From | Oleg Nesterov <> | Subject | Re: [PATCH 2/2] ptrace: fix set_task_blockstep()->update_debugctlmsr() logic |
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Hi.
Today I noticed by accident that starting from Aug 4 (at least) all my emails went to nowhere. I am resending some of them...
On 08/07, Sebastian Andrzej Siewior wrote: > > On 08/03/2012 06:29 PM, Oleg Nesterov wrote: >> --- a/arch/x86/kernel/step.c >> +++ b/arch/x86/kernel/step.c >> @@ -166,12 +166,18 @@ static void set_task_blockstep(struct task_struct *task, bool on) >> else >> clear_tsk_thread_flag(task, TIF_BLOCKSTEP); >> >> + if (task != current) >> + return; >> + >> + /* ensure irq/preemption can't change debugctl in between */ >> + local_irq_disable(); >> debugctl = get_debugctlmsr(); >> if (on) >> debugctl |= DEBUGCTLMSR_BTF; >> else >> debugctl&= ~DEBUGCTLMSR_BTF; >> update_debugctlmsr(debugctl); >> + local_irq_enable(); >> } > > I would say that you can remove this chunk. For task != current we > leave.
It turns out, original code is even more buggy than I thought.
Ironically, "task != current" case is more difficult and so far I do not see how we can handle this case correctly. I'll return to this a bit later, currently I am working on other patches.
> For uprobes we never set the bit, we only need it cleared.
Yes, at least at first step, and probably we will never need more.
> We get here > via int 3 and do_debug() already clears TIF_BLOCKSTEP
No, we get here via do_int3(), TIF_BLOCKSTEP is not cleared,
> because the > CPU clears the bit in CPU.
I am not sure. The manual says:
If the BTF flag is set when the processor generates a debug exception, the processor clears the BTF flag along with the TF flag.
but I am not sure "debug exception" also means "breakpoint exception".
do_debug() does clear TIF_BLOCKSTEP, and "The processor cleared BTF" is true in this case. But it is called after single-step.
Oleg.
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