Re: [Qemu-devel] [PATCH] User mode: Handle x86_64 vsyscall

[Richard Henderson <rth@twiddle.net>]

On 02/06/2010 03:50 PM, Laurent Desnogues wrote:
>> * target-i386 code should not have to know about
>>   linux vsyscall
>
> Given that we have to workaround 64-bit virtual
> address limitations (cf. Richard mail and previous
> discussions on the list), doing otherwise looks
> difficult.

Actually, it should be easy for QEMU to handle this.

The application is given the address of the VDSO in the AT_SYSINFO and 
AT_SYSINFO_EHDR entries of the auxvec (on the stack above argv and 
environ).  We can place this anywhere we like; the fact that the kernel 
puts it in high memory is merely a convenience to the kernel.

There *is* a legacy vsyscall address in high memory, from before the 
whole VDSO arrangement was worked out, but we could probably get away 
with ignoring that.  Certainly well behaved applications will be 
honoring the VDSO when it is given.

>> * it is not possible to step into vsyscall code
>>   using a debugger
>
> How would you achieve that?  Your guest OS
> doesn't necessarily have the code mapped.  I
> think this has to be considered as other syscalls,
> though slightly different.

If QEMU implements the VDSO, the page *will* be mapped, and the debugger 
will Just Work.

I imagine that QEMU's VDSO would not have the complicated bits that the 
kernel's version does, where it arranges to read the clock without going 
into kernel space.  I imagine QEMU would simply stuff a normal syscall 
sequence in there, which would automatically be emulated in the normal way.

Have a stare at the linux/arch/x86/vdso directory to see how things work.


r~