Messages in this thread | | | Date | Tue, 07 Aug 2012 17:29:38 +0200 | From | Sebastian Andrzej Siewior <> | Subject | Re: [PATCH 2/2] ptrace: fix set_task_blockstep()->update_debugctlmsr() logic |
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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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