Re: [Valgrind-users] Does using modify_ldt break Valgrind?

[Alex B. <ker...@be...>]

On Wed, 2005-11-02 at 11:20 +0000, Tom Hughes wrote:
> In message <113...@ok...>
>         Alex Bennee <ker...@be...> wrote:
> 
> > I'm trying to run Valgrind across an x86 app that uses the modify_ldt
> > syscall which seems to get in the way of Valgrind. 
> > <snip>
> 
> It's not that it is used by valgrind, it is virtualised by valgrind
> because it needs to alter the LDT for the virtual CPU that your code
> is running on.
>
> <snip>
>
> Changing VEX_GUEST_X86_LDT_NENT in VEX/pub/libvex_guest_x86.h to
> a larger number (8192 to get the full LDT) should allow your code
> to work I think.

Thanks. That worked and it moved forward. Unfortunately we set up the fs
segment selector to point at the entry. This doesn't seem to be
correctly setup under valgrind:

setFS(0x3ef)
checkFs(0x427ae40)
==14078==
==14078== Process terminating with default action of signal 11
(SIGSEGV): dumping core
==14078==  General Protection Fault
==14078==    at 0x3808BA9D: peekFs(int) (SegmentSelector.cc:88)
<snip>
==14078== starting debugger with cmd: /usr/bin/gdb 
<snip>
0x3808ba9d in peekFs (offset=0) at SegmentSelector.cc:88
88        __asm__ volatile(
(gdb) p $fs
$1 = -1342177280
(gdb) p/x
$2 = 0xb0000000

Vs running direct under gdb:

<snip>
Breakpoint 1, peekFs (offset=0) at SegmentSelector.cc:88
88        __asm__ volatile(
(gdb) p $fs
$1 = 1007

> 
> Tom

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