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: NIngo Molnar <mingo@redhat.com>
Cc: Thomas Gleixner <tglx@timesys.com>
Signed-off-by: NChristian Borntraeger <borntrae@de.ibm.com>
Signed-off-by: NAndrew Morton <akpm@osdl.org>
Signed-off-by: NLinus Torvalds <torvalds@osdl.org>