Re: [ltt-dev] [RFC git tree] Userspace RCU (urcu) for Linux (repost)

["Paul E. McKenney" <paulmck@linux.vnet.ibm.com>]

On Mon, Feb 09, 2009 at 10:37:42AM -0800, Paul E. McKenney wrote:
> On Mon, Feb 09, 2009 at 01:13:41PM -0500, Mathieu Desnoyers wrote:
> > * Paul E. McKenney (paulmck@linux.vnet.ibm.com) wrote:

[ . . . ]

> > You know what ? Changing RCU_GP_CTR_BIT to 16 uses a
> > testw %ax, %ax instead of a testb %al, %al. The trick here is that
> > RCU_GP_CTR_BIT must be a multiple of 8 so we can use a full 8-bits,
> > 16-bits or 32-bits bitmask for the lower order bits.
> > 
> > On 64-bits, using a RCU_GP_CTR_BIT of 32 is also ok. It uses a testl.
> > 
> > To provide 32-bits compability and allow the deepest nesting possible, I
> > think it makes sense to use
> > 
> > /* Use the amount of bits equal to half of the architecture long size */
> > #define RCU_GP_CTR_BIT (sizeof(long) << 2)
> 
> You lost me on this one:
> 
> 	sizeof(long) << 2 = 0x10
> 
> I could believe the following (run on a 32-bit machine):
> 
> 	1 << (sizeof(long) * 8 - 1) = 0x80000000
> 
> Or, if you were wanting to use a bit halfway up the word, perhaps this:
> 
> 	1 << (sizeof(long) * 4 - 1) = 0x8000
> 
> Or am I confused?

Well, I am at least partly confused.  You were wanting a low-order bit,
so you want to lose the "- 1" above.  Here are some of the possibilities:

	sizeof(long) = 0x4
	sizeof(long) << 2 = 0x10
	1 << (sizeof(long) * 8 - 1) = 0x80000000
	1 << (sizeof(long) * 4) = 0x10000
	1 << (sizeof(long) * 4 - 1) = 0x8000
	1 << (sizeof(long) * 2) = 0x100
	1 << (sizeof(long) * 2 - 1) = 0x80

My guess is that 1 << (sizeof(long) * 4) and 1 << (sizeof(long) * 2)
are of the most interest.

							Thanx, Paul