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Message-ID: <286AC319A985734F985F78AFA26841F739376DA1@shsmsx102.ccr.corp.intel.com>
Date: Fri, 1 Dec 2017 15:09:08 +0000
From: "Wang, Wei W" <wei.w.wang@...el.com>
To: Tetsuo Handa <penguin-kernel@...ove.SAKURA.ne.jp>
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Subject: RE: [PATCH v18 05/10] xbitmap: add more operations
On Friday, December 1, 2017 9:02 PM, Tetsuo Handa wrote:
> Wei Wang wrote:
> > On 11/30/2017 06:34 PM, Tetsuo Handa wrote:
> > > Wei Wang wrote:
> > >> + * @start: the start of the bit range, inclusive
> > >> + * @end: the end of the bit range, inclusive
> > >> + *
> > >> + * This function is used to clear a bit in the xbitmap. If all the
> > >> +bits of the
> > >> + * bitmap are 0, the bitmap will be freed.
> > >> + */
> > >> +void xb_clear_bit_range(struct xb *xb, unsigned long start,
> > >> +unsigned long end) {
> > >> + struct radix_tree_root *root = &xb->xbrt;
> > >> + struct radix_tree_node *node;
> > >> + void **slot;
> > >> + struct ida_bitmap *bitmap;
> > >> + unsigned int nbits;
> > >> +
> > >> + for (; start < end; start = (start | (IDA_BITMAP_BITS - 1)) + 1) {
> > >> + unsigned long index = start / IDA_BITMAP_BITS;
> > >> + unsigned long bit = start % IDA_BITMAP_BITS;
> > >> +
> > >> + bitmap = __radix_tree_lookup(root, index, &node, &slot);
> > >> + if (radix_tree_exception(bitmap)) {
> > >> + unsigned long ebit = bit + 2;
> > >> + unsigned long tmp = (unsigned long)bitmap;
> > >> +
> > >> + nbits = min(end - start + 1, BITS_PER_LONG - ebit);
> > > "nbits = min(end - start + 1," seems to expect that start == end is
> > > legal for clearing only 1 bit. But this function is no-op if start == end.
> > > Please clarify what "inclusive" intended.
> >
> > If xb_clear_bit_range(xb,10,10), then it is effectively the same as
> > xb_clear_bit(10). Why would it be illegal?
> >
> > "@start inclusive" means that the @start will also be included to be
> > cleared.
>
> If start == end is legal,
>
> for (; start < end; start = (start | (IDA_BITMAP_BITS - 1)) + 1) {
>
> makes this loop do nothing because 10 < 10 is false.
How about "start <= end "?
Best,
Wei
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