Messages in this thread | | | Date | Mon, 30 Mar 2015 16:27:10 +0200 | From | Denys Vlasenko <> | Subject | Re: [PATCH] x86/asm/entry/64: better check for canonical address |
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On 03/26/2015 07:45 PM, Andy Lutomirski wrote: > On Thu, Mar 26, 2015 at 5:42 AM, Denys Vlasenko <dvlasenk@redhat.com> wrote: >> This change makes the check exact (no more false positives >> on kernel addresses). >> >> It isn't really important to be fully correct here - >> almost all addresses we'll ever see will be userspace ones, >> but OTOH it looks to be cheap enough: >> the new code uses two more ALU ops but preserves %rcx, >> allowing to not reload it from pt_regs->cx again. >> On disassembly level, the changes are: >> >> cmp %rcx,0x80(%rsp) -> mov 0x80(%rsp),%r11; cmp %rcx,%r11 >> shr $0x2f,%rcx -> shl $0x10,%rcx; sar $0x10,%rcx; cmp %rcx,%r11 >> mov 0x58(%rsp),%rcx -> (eliminated) >> >> Signed-off-by: Denys Vlasenko <dvlasenk@redhat.com> >> CC: Borislav Petkov <bp@alien8.de> >> CC: x86@kernel.org >> CC: linux-kernel@vger.kernel.org >> --- >> >> Andy, I'd undecided myself on the merits of doing this. >> If you like it, feel free to take it in your tree. >> I trimmed CC list to not bother too many people with this trivial >> and quite possibly "useless churn"-class change. > > I suspect that the two added ALU ops are free for all practical > purposes, and the performance of this path isn't *that* critical. > > If anyone is running with vsyscall=native because they need the > performance, then this would be a big win. Otherwise I don't have a > real preference. Anyone else have any thoughts here? > > Let me just run through the math quickly to make sure I believe all the numbers: > > Canonical addresses either start with 17 zeros or 17 ones. > > In the old code, we checked that the top (64-47) = 17 bits were all > zero. We did this by shifting right by 47 bits and making sure that > nothing was left. > > In the new code, we're shifting left by (64 - 48) = 16 bits and then > signed shifting right by the same amount, this propagating the 17th > highest bit to all positions to its left. If we get the same value we > started with, then we're good to go. > > So it looks okay to me.
So please take it into your tree :)
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