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metag: fix memory barriers
Volatile access doesn't really imply the compiler barrier. Volatile access is only ordered with respect to other volatile accesses, it isn't ordered with respect to general memory accesses. Gcc may reorder memory accesses around volatile access, as we can see in this simple example (if we compile it with optimization, both increments of *b will be collapsed to just one): void fn(volatile int *a, long *b) { (*b)++; *a = 10; (*b)++; } Consequently, we need the compiler barrier after a write to the volatile variable, to make sure that the compiler doesn't reorder the volatile write with something else. Signed-off-by: Mikulas Patocka <mpatocka@redhat.com> Cc: stable@vger.kernel.org Acked-by: Peter Zijlstra <peterz@infradead.org> Signed-off-by: James Hogan <james.hogan@imgtec.com>
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@ -15,6 +15,7 @@ static inline void wr_fence(void)
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volatile int *flushptr = (volatile int *) LINSYSEVENT_WR_FENCE;
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volatile int *flushptr = (volatile int *) LINSYSEVENT_WR_FENCE;
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barrier();
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barrier();
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*flushptr = 0;
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*flushptr = 0;
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barrier();
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}
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}
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#else /* CONFIG_METAG_META21 */
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#else /* CONFIG_METAG_META21 */
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@ -35,6 +36,7 @@ static inline void wr_fence(void)
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*flushptr = 0;
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*flushptr = 0;
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*flushptr = 0;
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*flushptr = 0;
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*flushptr = 0;
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*flushptr = 0;
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barrier();
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}
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}
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#endif /* !CONFIG_METAG_META21 */
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#endif /* !CONFIG_METAG_META21 */
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@ -68,6 +70,7 @@ static inline void fence(void)
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volatile int *flushptr = (volatile int *) LINSYSEVENT_WR_ATOMIC_UNLOCK;
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volatile int *flushptr = (volatile int *) LINSYSEVENT_WR_ATOMIC_UNLOCK;
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barrier();
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barrier();
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*flushptr = 0;
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*flushptr = 0;
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barrier();
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}
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}
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#define smp_mb() fence()
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#define smp_mb() fence()
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#define smp_rmb() fence()
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#define smp_rmb() fence()
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