[PATCH] bitmap: speedup in bitmap_find_free_region when order is 0

[PATCH] bitmap: speedup in bitmap_find_free_region when order is 0

[Chanho Min]

If bitmap_find_free_region() is called with order=0, We can reduce
for-loops to find 1 free bit. First, It scans bitmap array by the
increment of long type, then find 1 free bit within 1 long type value.

In 32 bits system and 1024 bits size, in the worst case, We need 1024
for-loops to find 1 free bit. But, If This is applied, it takes
64 for-loops. Instead, if free bit is in the first index of the bitmaps,
It will be needed additional 1 for-loop. But from second index, It
will be speed up significantly.

Signed-off-by: Chanho Min <chanho.min@lge.com>
---
 lib/bitmap.c |   34 ++++++++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)

diff --git a/lib/bitmap.c b/lib/bitmap.c
index 06f7e4f..0c1f03a 100644
--- a/lib/bitmap.c
+++ b/lib/bitmap.c
@@ -1099,6 +1099,37 @@ done:
 }
 
 /**
+ * bitmap_find_free_one - find a mem region
+ *	@bitmap: array of unsigned longs corresponding to the bitmap
+ *	@bits: number of bits in the bitmap
+ *
+ * Find one of free (zero) bits in a @bitmap of @bits bits and
+ * allocate them (set them to one).
+ *
+ * Return the bit offset in bitmap of the allocated region,
+ * or -errno on failure.
+ */
+static int __bitmap_find_free_one(unsigned long *bitmap, int bits)
+{
+	int pos, i;
+	unsigned long mask = (unsigned long)(~((unsigned long) 0));
+	int nlongs_reg = BITS_TO_LONGS(bits);
+
+	for (i = 0 ; i < nlongs_reg ; i++) {
+		if ((bitmap[i] & mask) != mask) {
+			for (pos = 0 ; pos < BITS_PER_LONG ; pos++) {
+				if (!__reg_op(&bitmap[i], pos, 0,
+					REG_OP_ISFREE))
+					continue;
+				__reg_op(&bitmap[i], pos, 0, REG_OP_ALLOC);
+				return pos;
+			}
+		}
+	}
+	return -ENOMEM;
+}
+
+/**
  * bitmap_find_free_region - find a contiguous aligned mem region
  *	@bitmap: array of unsigned longs corresponding to the bitmap
  *	@bits: number of bits in the bitmap
@@ -1116,6 +1147,9 @@ int bitmap_find_free_region(unsigned long *bitmap, int bits, int order)
 {
 	int pos, end;		/* scans bitmap by regions of size order */
 
+	if (order == 0)
+		return __bitmap_find_free_one(bitmap, bits);
+
 	for (pos = 0 ; (end = pos + (1 << order)) <= bits; pos = end) {
 		if (!__reg_op(bitmap, pos, order, REG_OP_ISFREE))
 			continue;
-- 
1.7.9.5

Re: [PATCH] bitmap: speedup in bitmap_find_free_region when order is 0

[Andrew Morton]

On Mon,  8 Apr 2013 11:23:48 +0900 Chanho Min <chanho.min@lge.com> wrote:

> If bitmap_find_free_region() is called with order=0, We can reduce
> for-loops to find 1 free bit. First, It scans bitmap array by the
> increment of long type, then find 1 free bit within 1 long type value.

That seems sensible.  I assume-without-checking that single-bit is a
common case?

> In 32 bits system and 1024 bits size, in the worst case, We need 1024
> for-loops to find 1 free bit. But, If This is applied, it takes
> 64 for-loops. Instead, if free bit is in the first index of the bitmaps,
> It will be needed additional 1 for-loop. But from second index, It
> will be speed up significantly.
> 
> --- a/lib/bitmap.c
> +++ b/lib/bitmap.c
> @@ -1099,6 +1099,37 @@ done:
>  }
>  
>  /**
> + * bitmap_find_free_one - find a mem region
> + *	@bitmap: array of unsigned longs corresponding to the bitmap
> + *	@bits: number of bits in the bitmap
> + *
> + * Find one of free (zero) bits in a @bitmap of @bits bits and
> + * allocate them (set them to one).
> + *
> + * Return the bit offset in bitmap of the allocated region,
> + * or -errno on failure.
> + */
> +static int __bitmap_find_free_one(unsigned long *bitmap, int bits)
> +{
> +	int pos, i;
> +	unsigned long mask = (unsigned long)(~((unsigned long) 0));

That seems unnecessarily complicated.  "unsigned long mask = -1;" works :)

> +	int nlongs_reg = BITS_TO_LONGS(bits);
> +
> +	for (i = 0 ; i < nlongs_reg ; i++) {
> +		if ((bitmap[i] & mask) != mask) {

But here we could just do "if (bitmap[i] != -1)".  Or ~0UL.

> +			for (pos = 0 ; pos < BITS_PER_LONG ; pos++) {
> +				if (!__reg_op(&bitmap[i], pos, 0,
> +					REG_OP_ISFREE))
> +					continue;
> +				__reg_op(&bitmap[i], pos, 0, REG_OP_ALLOC);
> +				return pos;
> +			}
> +		}
> +	}
> +	return -ENOMEM;
> +}

afacit the code is buggy - if `bits' is not an exact multiple of
BITS_PER_LONG, this search will wander off the end of the specified
region?

> +/**
>   * bitmap_find_free_region - find a contiguous aligned mem region
>   *	@bitmap: array of unsigned longs corresponding to the bitmap
>   *	@bits: number of bits in the bitmap
> @@ -1116,6 +1147,9 @@ int bitmap_find_free_region(unsigned long *bitmap, int bits, int order)
>  {
>  	int pos, end;		/* scans bitmap by regions of size order */
>  
> +	if (order == 0)
> +		return __bitmap_find_free_one(bitmap, bits);
> +
>  	for (pos = 0 ; (end = pos + (1 << order)) <= bits; pos = end) {
>  		if (!__reg_op(bitmap, pos, order, REG_OP_ISFREE))
>  			continue;

Re: [PATCH] bitmap: speedup in bitmap_find_free_region when order is 0

[Joe Perches]

On Mon, 2013-04-08 at 14:30 -0700, Andrew Morton wrote:
> On Mon,  8 Apr 2013 11:23:48 +0900 Chanho Min <chanho.min@lge.com> wrote:
> > --- a/lib/bitmap.c
[]
> > +static int __bitmap_find_free_one(unsigned long *bitmap, int bits)
[]
> > +	unsigned long mask = (unsigned long)(~((unsigned long) 0));
> 
> That seems unnecessarily complicated.  "unsigned long mask = -1;" works :)

So would ULONG_MAX or just ~0UL