On Thu, Apr 02, 2015 at 11:57:59AM -0400, John Snow wrote:
> 
> 
> On 04/02/2015 09:37 AM, Stefan Hajnoczi wrote:
> >On Fri, Mar 20, 2015 at 03:16:58PM -0400, John Snow wrote:
> >>+void hbitmap_truncate(HBitmap *hb, uint64_t size)
> >>+{
> >>+    bool shrink;
> >>+    unsigned i;
> >>+    uint64_t num_elements = size;
> >>+    uint64_t old;
> >>+
> >>+    /* Size comes in as logical elements, adjust for granularity. */
> >>+    size = (size + (1ULL << hb->granularity) - 1) >> hb->granularity;
> >>+    assert(size <= ((uint64_t)1 << HBITMAP_LOG_MAX_SIZE));
> >>+    shrink = size < hb->size;
> >>+
> >>+    /* bit sizes are identical; nothing to do. */
> >>+    if (size == hb->size) {
> >>+        return;
> >>+    }
> >>+
> >>+    /* If we're losing bits, let's clear those bits before we invalidate all of
> >>+     * our invariants. This helps keep the bitcount consistent, and will prevent
> >>+     * us from carrying around garbage bits beyond the end of the map.
> >>+     *
> >>+     * Because clearing bits past the end of map might reset bits we care about
> >>+     * within the array, record the current value of the last bit we're keeping.
> >>+     */
> >>+    if (shrink) {
> >>+        bool set = hbitmap_get(hb, num_elements - 1);
> >>+        uint64_t fix_count = (hb->size << hb->granularity) - num_elements;
> >>+
> >>+        assert(fix_count);
> >>+        hbitmap_reset(hb, num_elements, fix_count);
> >>+        if (set) {
> >>+            hbitmap_set(hb, num_elements - 1, 1);
> >>+        }
> >
> >Why is it necessary to set the last bit (if it was set)?  The comment
> >isn't clear to me.
> >
> 
> Sure. The granularity of the bitmap provides us with virtual bit groups. for
> a granularity of say g=2, we have 2^2 virtual bits per every real bit:
> 
> 101 in memory is treated, virtually, as 1111 0000 1111.
> 
> The get/set calls operate on virtual bits, not concrete ones, so if we were
> to reset virtual bits 2-11:
> 11|11 0000 1111
> 
> We'd set the real bits to '000', because we clear or set the entire virtual
> group.
> 
> This is probably not what we really want, so as a shortcut I just read and
> then re-set the final bit.
> 
> It is programmatically avoidable (Are we truncating into a granularity
> group?) but in the case that we are, I'd need to read/reset the bit anyway,
> so it seemed fine to just unconditionally apply the fix.

I see.  This is equivalent to:

uint64_t start = QEMU_ALIGN_UP(num_elements, hb->granularity);
uint64_t fix_count = (hb->size << hb->granularity) - start;
hbitmap_reset(hb, start, fix_count);

The explicit QEMU_ALIGN_UP(num_elements, hb->granularity) calculation
shows that we're working around the granularity.  I find this easier to
understand.

If you keep the get/set version, please extend the comment to explain
that clearing the first bit could destroy up to granularity - 1 bits
that must be preserved.

Stefan