On Mon, May 02, 2016 at 10:33:10AM -0700, Dave Hansen wrote:
> On 05/02/2016 10:19 AM, Yu-cheng Yu wrote:
> > On Mon, May 02, 2016 at 09:43:47AM -0700, Dave Hansen wrote:
> >>> If (fpu.fpstate_active == 0), then the task does not use FPU; we don't
> >>> want to save these registers, right?
> >>
> >> No. It's possible to have fpstate_active=0 while fpregs_active=1. Such
> >> a task uses the FPU, but just hasn't done an XSAVE* to save the register
> >> content to the fpstate buffer.
> >>
> >> Note, this is just theoretical, and does not happen in this particular
> >> call path today.
> >
> > What about...
> >
> > static int may_copy_fpregs_to_sigframe(void)
> > {
> > if (fpregs_active())
> > return 1;
> >
> > WARN_ONCE(!current->thread.fpu.fpstate_active,
> > "direct FPU save with no math use\n");
> >
> > if (boot_cpu_has(X86_FEATURE_XSAVES))
> > return 1;
> >
> > return 0;
> > }
>
> I don't think that changes anything. We still have a check in there
> that has no purpose. You've changed the ordering so that the specific
> example that I pointed out no longer triggers it. But, the underlying
> issue remains.
Before Linux gets into copy_fpstate_to_sigframe(),
current->thread.fpu.fpstate_active must be true.
For eagerfpu, fpregs_active() must also be true.
For lazyfpu, once we try to do FSAVE/FXSAVE/XSAVE,
fpregs_active() will become true as well.
We should have not based on boot_cpu_has(X86_FEATURE_XSAVES)
at all.
Why don't we make it simple and always copy_fpregs_to_signal_frame()?
Or, only for the lazy case, i.e. !fpregs_active(), we do __copy_to_user().
Anyway, I think we can just replace may_copy_fpregs_to_sigframe() with
!fpregs_active().
Comments?
