On Sun, 2005-12-18 at 19:27 +0200, Pekka Enberg wrote:
> Hi Steven,
>
> On 12/18/05, Steven Rostedt <[email protected]> wrote:
> > + do {
> > + x = 1;
> > + while ((x+i)*size + ALIGN(base+(x+i)*extra, align) <= wastage)
> > + x <<= 1;
> > + i += (x >> 1);
> > + } while (x > 1);
>
> The above is pretty hard to read. Perhaps we could give x and i better
> names? Also, couldn't we move left part of the expression into a
> separate static inline function for readability?
Actually, Luuk sent me this patch made by Balbir Singh that was done a
while ago.
extra = sizeof(kmem_bufctl_t);
}
- i = 0;
+ i = (wastage - base)/(size + extra);
while (i*size + L1_CACHE_ALIGN(base+i*extra) <= wastage)
i++;
- if (i > 0)
+ while (i*size + L1_CACHE_ALIGN(base+i*extra) > wastage)
i--;
This actually has a O(1) with a K=2. Analyzing this further, I've come
up with the below patch. This patch removes the need for the second
while, and adds a comment to why. The size is already calculated to be
no smaller than the alignment. So that the division will not return
something greater than 1 of what is needed. So the if (i > 0) after the
while is all that is needed. So this patch is O(1) K=1.
-- Steve
Index: linux-2.6.15-rc5/mm/slab.c
===================================================================
--- linux-2.6.15-rc5.orig/mm/slab.c 2005-12-16 16:24:09.000000000 -0500
+++ linux-2.6.15-rc5/mm/slab.c 2005-12-18 13:30:13.000000000 -0500
@@ -708,7 +708,14 @@
base = sizeof(struct slab);
extra = sizeof(kmem_bufctl_t);
}
- i = 0;
+ /*
+ * Divide the amount we have, by the amount we need for
+ * each object. Since the size is already calculated
+ * to be no less than the alignment, this result will
+ * not be any greater than 1 that we need, and this will
+ * be subtracted after the while loop.
+ */
+ i = (wastage - base)/(size + extra);
while (i*size + ALIGN(base+i*extra, align) <= wastage)
i++;
if (i > 0)
-
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