of p greater
than or equal to v.  Its value is computed something like this:
	((v - 1) | (p - 1)) + 1
Let's consider using this in the context of the 3 cases above.

When p = 2^0 = 1, the result boils down to ((v - 1) | 0) + 1, so it
just translates any value v to itself.  That handles case (1) above.

When p = 2^n, n > 0, we know that (p - 1) will be a mask with all n
bits 0..n-1 set.  The condition in this case occurs when none of
those mask bits is set in the value v provided.  If that is the
case, subtracting 1 from v will have 1's in all those lower bits (at
least).  Therefore, OR-ing the mask with that decremented value has
no effect, so adding the 1 back again will just translate the v to
itself.  This handles case (2).

Otherwise, the value v is greater than some multiple of p, and
decrementing it will produce a result greater than or equal to that
multiple.  OR-ing in the mask will produce a value 1 less than the
next multiple of p, so finally adding 1 back will result in the
desired rounded-up value.  This handles case (3).

Hopefully this is convincing.

While I was at it, I converted XLOG_SECTOR_ROUNDDOWN_BLKNO() to use
the round_down() macro.

Signed-off-by: Alex Elder <aelder@sgi.com>
Reviewed-by: Christoph Hellwig <hch@lst.de>
Signed-off-by: Dave Chinner <dchinner@redhat.com>

é3