Re: [PATCH v9 05/19] qspinlock: Optimize for smaller NR_CPUS

[Waiman Long]

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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