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Message-ID: <50213462.5040304@linutronix.de>
Date: Tue, 07 Aug 2012 17:29:38 +0200
From: Sebastian Andrzej Siewior <bigeasy@...utronix.de>
To: Oleg Nesterov <oleg@...hat.com>
CC: Ingo Molnar <mingo@...e.hu>,
Ananth N Mavinakayanahalli <ananth@...ibm.com>,
Anton Arapov <anton@...hat.com>,
"H. Peter Anvin" <hpa@...or.com>,
Peter Zijlstra <peterz@...radead.org>,
Roland McGrath <roland@...k.frob.com>,
Srikar Dronamraju <srikar@...ux.vnet.ibm.com>,
linux-kernel@...r.kernel.org
Subject: Re: [PATCH 2/2] ptrace: fix set_task_blockstep()->update_debugctlmsr()
logic
On 08/07/2012 05:15 PM, Oleg Nesterov wrote:
> 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.
maybe you could remove the autodectect mode and add helper for uprobe
which disables it.
>> 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,
Yes, Sorry. my fault.
>> 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.
I was wrong here in regard to do_debug() since do_int3() is correct.
Anyway, I checked it on real hardware and I saw the CPU in do_int3()
with BTF set after executing int3 with TF flag set and the BTF bit.
>
> Oleg.
Sebastian
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