On 04/18/2014 04:27 AM, Peter Zijlstra wrote:
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].
I am confused by your notation. Anyway, my version of the byte location
chart is:
val is LE bytes[0,1,2,3] BE [0,1,2,3]
locked_pending is LE bytes[0,1] BE [2,3]
locked is LE bytes[0] BE [3]
If we assign 1 to BE locked_pending, bytes[2] = 0 and bytes[3] = 1. Note
that the LSB of the BE locked_pending is bytes[3]. Similarly, if we
assign 1 to BE val, bytes[3] = 1 and all the other bytes will be 0.
-Longman
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