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

[Andy Lutomirski]

On Tue, Oct 27, 2020 at 1:19 AM Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>
> On Mon, Oct 26, 2020 at 04:30:32PM -0700, Andy Lutomirski wrote:
>
> > > @@ -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.
>
> Kyle also mentioned it. The reason I didn't do that is because ptrace()
> didn't set the TF, so why should it see it in ptrace_get_debugreg(6) ?

I assume you already figured this out, but my specific concern is with
the get_si_code(dr6) part -- that's sent directly to the task being
debugged or debugging itself (and, sadly, to ptrace, and who knows
what debuggers do).