[Valgrind-users] Does using modify_ldt break Valgrind?

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

Hi,

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. I has a look though
the svn sources and I can see the test case for modify_ldt is commented
out and has a note that it is used by Valgrind itself. Running outside
Valgrind gives:

$: ./mytestprog
ldt_entry.entry_number   = 125
ldt_entry.base_addr      = 3821ec00
ldt_entry.limit          = 9944
ldt_entry.seg_32bit      = 1
ldt_entry.contents       = 0
ldt_entry.read_exec_only = 0
ldt_entry.limit_in_pages = 0
ldt_entry.seg_not_present= 0
ldt_entry.useable        = 1
modify_ldt = 0
setFS(1007)
libc_nbtest: starting
<snip>

Whereas under Valgrind:
$ /usr/local/bin/valgrind --tool=memcheck ./mytestprog
==9588== Memcheck, a memory error detector.
==9588== Copyright (C) 2002-2005, and GNU GPL'd, by Julian Seward et al.
==9588== Using LibVEX rev 1426, a library for dynamic binary
translation.
==9588== Copyright (C) 2004-2005, and GNU GPL'd, by OpenWorks LLP.
==9588== Using valgrind-3.1.SVN, a dynamic binary instrumentation
framework.
==9588== Copyright (C) 2000-2005, and GNU GPL'd, by Julian Seward et al.
==9588== For more details, rerun with: -v
==9588==
ldt_entry.entry_number   = 125
ldt_entry.base_addr      = 427ae40
ldt_entry.limit          = 9944
ldt_entry.seg_32bit      = 1
ldt_entry.contents       = 0
ldt_entry.read_exec_only = 0
ldt_entry.limit_in_pages = 0
ldt_entry.seg_not_present= 0
ldt_entry.useable        = 1
modify_ldt = -1
FATAL ERROR: User segment selector installation failed.
  Are user selectors enabled?

I'm assuming the move in base_addr is because memory has been moved
about a bit. I've tried with both the formal 3.0.1 and the SVN sources.
Is my attempt to Valgrind this app doomed because of use of the
modify_ldt?


--
Alex, homepage: http://www.bennee.com/~alex/
You might have mail