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SubjectRe: [PATCHv3 5/8] x86/uaccess: Provide untagged_addr() and remove tags before address check
On Thu, Jun 16, 2022 at 11:30:49AM +0200, Peter Zijlstra wrote:
> On Mon, Jun 13, 2022 at 05:36:43PM +0000, Edgecombe, Rick P wrote:
> > On Fri, 2022-06-10 at 17:35 +0300, Kirill A. Shutemov wrote:
> > > +#ifdef CONFIG_X86_64
> > > +/*
> > > + * Mask out tag bits from the address.
> > > + *
> > > + * Magic with the 'sign' allows to untag userspace pointer without
> > > any branches
> > > + * while leaving kernel addresses intact.
> >
> > Trying to understand the magic part here. I guess how it works is, when
> > the high bit is set, it does the opposite of untagging the addresses by
> > setting the tag bits instead of clearing them. So:
>
> The magic is really rather simple to see; there's two observations:
>
> x ^ y ^ y == x
>
> That is; xor is it's own inverse. And secondly, xor with 1 is a bit
> toggle.
>
> So if we mask a negative value, we destroy the sign. Therefore, if we
> xor with the sign-bit, we have a nop for positive numbers and a toggle
> for negatives (effectively making them positive, -1, 2s complement
> yada-yada) then we can mask, without fear of destroying the sign, and
> then we xor again to undo whatever we did before, effectively restoring
> the sign.
>
> Anyway, concequence of all this is that LAM_U48 won't work correct on
> 5-level kernels, because the mask will still destroy kernel pointers.

Any objection against this variant (was posted in the thread):

#define untagged_addr(mm, addr) ({ \
u64 __addr = (__force u64)(addr); \
s64 sign = (s64)__addr >> 63; \
__addr &= (mm)->context.untag_mask | sign; \
(__force __typeof__(addr))__addr; \
})

?

I find it easier to follow and it is LAM_U48-safe.

--
Kirill A. Shutemov

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