On Mon, 8 Apr 2013 11:23:48 +0900 Chanho Min <chanho....@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;
--
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