Re: [REGRESSION] x86/debug: After PTRACE_SINGLESTEP DR_STEP is no longer reported in dr6

[Andy Lutomirski]

On Mon, Oct 26, 2020 at 9:55 AM Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>
> On Mon, Oct 26, 2020 at 05:31:00PM +0100, Peter Zijlstra wrote:
> > In that respect, I think the current virtual_dr6 = 0 is placed wrong, it
> > should only be in exc_debug_user(). The only 'problem' then is that we
> > seem to be able to loose BTF, but perhaps that is already an extant bug.
> >
> > Consider:
> >
> >  - perf: setup in-kernel #DB
> >  - tracer: ptrace(PTRACE_SINGLEBLOCK)
> >  - tracee: #DB on perf breakpoint, looses BTF
> >  - tracee .. never triggers actual blockstep
> >
> > Hmm ? Should we re-set BTF when TIF_BLOCKSTEP && !user_mode(regs) ?
>
> Something like so then.
>
> Or sould we also have the userspace #DB re-set BTF when it was !DR_STEP?
> I need to go untangle that ptrace stuff :/
>
> diff --git a/arch/x86/kernel/traps.c b/arch/x86/kernel/traps.c
> index 3c70fb34028b..31de8b0980ca 100644
> --- a/arch/x86/kernel/traps.c
> +++ b/arch/x86/kernel/traps.c
> @@ -793,19 +793,6 @@ static __always_inline unsigned long debug_read_clear_dr6(void)
>         set_debugreg(DR6_RESERVED, 6);
>         dr6 ^= DR6_RESERVED; /* Flip to positive polarity */
>
> -       /*
> -        * Clear the virtual DR6 value, ptrace routines will set bits here for
> -        * things we want signals for.
> -        */
> -       current->thread.virtual_dr6 = 0;
> -
> -       /*
> -        * The SDM says "The processor clears the BTF flag when it
> -        * generates a debug exception."  Clear TIF_BLOCKSTEP to keep
> -        * TIF_BLOCKSTEP in sync with the hardware BTF flag.
> -        */
> -       clear_thread_flag(TIF_BLOCKSTEP);
> -
>         return dr6;
>  }
>
> @@ -873,6 +860,20 @@ static __always_inline void exc_debug_kernel(struct pt_regs *regs,
>          */
>         WARN_ON_ONCE(user_mode(regs));
>
> +       if (test_thread_flag(TIF_BLOCKSTEP)) {
> +               /*
> +                * The SDM says "The processor clears the BTF flag when it
> +                * generates a debug exception." but PTRACE_BLOCKSTEP requested
> +                * it for userspace, but we just took a kernel #DB, so re-set
> +                * BTF.
> +                */
> +               unsigned long debugctl;
> +
> +               rdmsrl(MSR_IA32_DEBUGCTLMSR, debugctl);
> +               debugctl |= DEBUGCTLMSR_BTF;
> +               wrmsrl(MSR_IA32_DEBUGCTLMSR, debugctl);
> +       }
> +
>         /*
>          * Catch SYSENTER with TF set and clear DR_STEP. If this hit a
>          * watchpoint at the same time then that will still be handled.
> @@ -935,6 +936,26 @@ static __always_inline void exc_debug_user(struct pt_regs *regs,
>         irqentry_enter_from_user_mode(regs);
>         instrumentation_begin();
>
> +       /*
> +        * Clear the virtual DR6 value, ptrace routines will set bits here for
> +        * things we want signals for.
> +        */
> +       current->thread.virtual_dr6 = 0;
> +
> +       /*
> +        * If PTRACE requested SINGLE(BLOCK)STEP, make sure to reflect that in
> +        * the ptrace visible DR6 copy.
> +        */
> +       if (test_thread_flag(TIF_BLOCKSTEP) || test_thread_flag(TIF_SINGLESTEP))
> +               current->thread.virtual_dr6 |= (dr6 & DR_STEP);

I'm guessing that this would fail a much simpler test, though: have a
program use PUSHF to set TF and then read out DR6 from the SIGTRAP.  I
can whip up such a test if you like.

Is there any compelling reason not to just drop the condition and do:

current->thread.virtual_dr6 |= (dr6 & DR_STEP);

unconditionally?  This DR6 cause, along with ICEBP, have the
regrettable distinctions of being the only causes that a user program
can trigger all on its own without informing the kernel first.  This
means that we can't fully separate the concept of "user mode is
single-stepping itself" from "ptrace or something else is causing the
kernel to single step a program."

I bet that, without making this tweak, the virtual_dr6 change will
regress some horrific Wine use case.

--Andy