struct reftable_record *rec)
{
struct pq_entry entry = { 0 };
- int err = 0;
+ int err = 0, empty;
+
+ empty = merged_iter_pqueue_is_empty(mi->pq);
if (mi->advance_index >= 0) {
+ /*
+ * When there are no pqueue entries then we only have a single
+ * subiter left. There is no need to use the pqueue in that
+ * case anymore as we know that the subiter will return entries
+ * in the correct order already.
+ *
+ * While this may sound like a very specific edge case, it may
+ * happen more frequently than you think. Most repositories
+ * will end up having a single large base table that contains
+ * most of the refs. It's thus likely that we exhaust all
+ * subiters but the one from that base ref.
+ */
+ if (empty)
+ return iterator_next(&mi->subiters[mi->advance_index].iter,
+ rec);
+
err = merged_iter_advance_subiter(mi, mi->advance_index);
if (err < 0)
return err;
+ if (!err)
+ empty = 0;
mi->advance_index = -1;
}
- if (merged_iter_pqueue_is_empty(mi->pq))
+ if (empty)
return 1;
entry = merged_iter_pqueue_remove(&mi->pq);