This patch adds a barrier() in futex unqueue_me to avoid aliasing of two
pointers.
On my s390x system I saw the following oops:
Unable to handle kernel pointer dereference at virtual kernel address
0000000000000000
Oops: 0004 [#1]
CPU: 0 Not tainted
Process mytool (pid: 13613, task: 000000003ecb6ac0, ksp: 00000000366bdbd8)
Krnl PSW : 0704d00180000000 00000000003c9ac2 (_spin_lock+0xe/0x30)
Krnl GPRS: 00000000ffffffff 000000003ecb6ac0 0000000000000000 0700000000000000
0000000000000000 0000000000000000 000001fe00002028 00000000000c091f
000001fe00002054 000001fe00002054 0000000000000000 00000000366bddc0
00000000005ef8c0 00000000003d00e8 0000000000144f91 00000000366bdcb8
Krnl Code: ba 4e 20 00 12 44 b9 16 00 3e a7 84 00 08 e3 e0 f0 88 00 04
Call Trace:
([<0000000000144f90>] unqueue_me+0x40/0xe4)
[<0000000000145a0c>] do_futex+0x33c/0xc40
[<000000000014643e>] sys_futex+0x12e/0x144
[<000000000010bb00>] sysc_noemu+0x10/0x16
[<000002000003741c>] 0x2000003741c
The code in question is:
static int unqueue_me(struct futex_q *q)
{
int ret = 0;
spinlock_t *lock_ptr;
/* In the common case we don't take the spinlock, which is nice. */
retry:
lock_ptr = q->lock_ptr;
if (lock_ptr != 0) {
spin_lock(lock_ptr);
/*
* q->lock_ptr can change between reading it and
* spin_lock(), causing us to take the wrong lock. This
* corrects the race condition.
[...]
and my compiler (gcc 4.1.0) makes the following out of it:
00000000000003c8 <unqueue_me>:
3c8: eb bf f0 70 00 24 stmg %r11,%r15,112(%r15)
3ce: c0 d0 00 00 00 00 larl %r13,3ce <unqueue_me+0x6>
3d0: R_390_PC32DBL .rodata+0x2a
3d4: a7 f1 1e 00 tml %r15,7680
3d8: a7 84 00 01 je 3da <unqueue_me+0x12>
3dc: b9 04 00 ef lgr %r14,%r15
3e0: a7 fb ff d0 aghi %r15,-48
3e4: b9 04 00 b2 lgr %r11,%r2
3e8: e3 e0 f0 98 00 24 stg %r14,152(%r15)
3ee: e3 c0 b0 28 00 04 lg %r12,40(%r11)
/* write q->lock_ptr in r12 */
3f4: b9 02 00 cc ltgr %r12,%r12
3f8: a7 84 00 4b je 48e <unqueue_me+0xc6>
/* if r12 is zero then jump over the code.... */
3fc: e3 20 b0 28 00 04 lg %r2,40(%r11)
/* write q->lock_ptr in r2 */
402: c0 e5 00 00 00 00 brasl %r14,402 <unqueue_me+0x3a>
404: R_390_PC32DBL _spin_lock+0x2
/* use r2 as parameter for spin_lock */
So the code becomes more or less:
if (q->lock_ptr != 0) spin_lock(q->lock_ptr)
instead of
if (lock_ptr != 0) spin_lock(lock_ptr)
Which caused the oops from above.
After adding a barrier gcc creates code without this problem:
[...] (the same)
3ee: e3 c0 b0 28 00 04 lg %r12,40(%r11)
3f4: b9 02 00 cc ltgr %r12,%r12
3f8: b9 04 00 2c lgr %r2,%r12
3fc: a7 84 00 48 je 48c <unqueue_me+0xc4>
400: c0 e5 00 00 00 00 brasl %r14,400 <unqueue_me+0x38>
402: R_390_PC32DBL _spin_lock+0x2
As a general note, this code of unqueue_me seems a bit fishy. The retry logic
of unqueue_me only works if we can guarantee, that the original value of
q->lock_ptr is always a spinlock (Otherwise we overwrite kernel memory). We
know that q->lock_ptr can change. I dont know what happens with the original
spinlock, as I am not an expert with the futex code.
Cc: Martin Schwidefsky <schwidefsky@de.ibm.com>
Cc: Rusty Russell <rusty@rustcorp.com.au>
Acked-by: Ingo Molnar <mingo@redhat.com>
Cc: Thomas Gleixner <tglx@timesys.com>
Signed-off-by: Christian Borntraeger <borntrae@de.ibm.com>
Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>
e91467ecd1