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.