Re: [PATCH 2/2] ptrace: fixset_task_blockstep()->update_debugctlmsr() logic
From: Oleg Nesterov
Date: Tue Aug 07 2012 - 11:18:56 EST
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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