On Thu, Apr 17, 2014 at 05:46:27PM -0400, Waiman Long wrote:
> On 04/17/2014 11:56 AM, Peter Zijlstra wrote:
> >On Thu, Apr 17, 2014 at 11:03:57AM -0400, Waiman Long wrote:
> >>+struct __qspinlock {
> >>+ union {
> >>+ atomic_t val;
char bytes[4];
> >>+ struct {
> >>+#ifdef __LITTLE_ENDIAN
> >>+ u16 locked_pending;
> >>+ u16 tail;
> >>+#else
> >>+ u16 tail;
> >>+ u16 locked_pending;
> >>+#endif
> >>+ };
struct {
#ifdef __LITTLE_ENDIAN
u8 locked;
#else
u8 res[3];
u8 locked;
#endif
};
> >>+ };
> >>+};
> >>+
> >>+/**
> >>+ * clear_pending_set_locked - take ownership and clear the pending bit.
> >>+ * @lock: Pointer to queue spinlock structure
> >>+ * @val : Current value of the queue spinlock 32-bit word
> >>+ *
> >>+ * *,1,0 -> *,0,1
> >>+ */
> >>+static __always_inline void
> >>+clear_pending_set_locked(struct qspinlock *lock, u32 val)
> >>+{
> >>+ struct __qspinlock *l = (void *)lock;
> >>+
> >>+ ACCESS_ONCE(l->locked_pending) = 1;
> >You lost the __constant_le16_to_cpu(_Q_LOCKED_VAL) there. The
> >unconditional 1 is wrong. You also have to flip the bytes in
> >locked_pending.
>
> I don't think that is wrong. The lock byte is in the least significant 8
> bits and the pending byte is the next higher significant 8 bits irrespective
> of the endian-ness. So a value of 1 in a 16-bit context means the lock byte
> is set, but the pending byte is cleared. The name "locked_pending" doesn't
> mean that locked variable is in a lower address than pending.
val is LE bytes[0,1,2,3] BE [3,2,1,0]
locked_pending is LE bytes[0,1] BE [1,0]
locked LE bytes[0] BE [0]
That does mean that the LSB of BE locked_pending is bytes[1].
So if you do BE: locked_pending = 1, you set bytes[1], not bytes[0].
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