static struct bfq_queue *
bfq_setup_cooperator(struct bfq_data *bfqd, struct bfq_queue *bfqq,
		     void *io_struct, bool request)
{
	struct bfq_queue *in_service_bfqq, *new_bfqq;

	/*
	 * Do not perform queue merging if the device is non
	 * rotational and performs internal queueing. In fact, such a
	 * device reaches a high speed through internal parallelism
	 * and pipelining. This means that, to reach a high
	 * throughput, it must have many requests enqueued at the same
	 * time. But, in this configuration, the internal scheduling
	 * algorithm of the device does exactly the job of queue
	 * merging: it reorders requests so as to obtain as much as
	 * possible a sequential I/O pattern. As a consequence, with
	 * the workload generated by processes doing interleaved I/O,
	 * the throughput reached by the device is likely to be the
	 * same, with and without queue merging.
	 *
	 * Disabling merging also provides a remarkable benefit in
	 * terms of throughput. Merging tends to make many workloads
	 * artificially more uneven, because of shared queues
	 * remaining non empty for incomparably more time than
	 * non-merged queues. This may accentuate workload
	 * asymmetries. For example, if one of the queues in a set of
	 * merged queues has a higher weight than a normal queue, then
	 * the shared queue may inherit such a high weight and, by
	 * staying almost always active, may force BFQ to perform I/O
	 * plugging most of the time. This evidently makes it harder
	 * for BFQ to let the device reach a high throughput.
	 *
	 * Finally, the likely() macro below is not used because one
	 * of the two branches is more likely than the other, but to
	 * have the code path after the following if() executed as
	 * fast as possible for the case of a non rotational device
	 * with queueing. We want it because this is the fastest kind
	 * of device. On the opposite end, the likely() may lengthen
	 * the execution time of BFQ for the case of slower devices
	 * (rotational or at least without queueing). But in this case
	 * the execution time of BFQ matters very little, if not at
	 * all.
	 */
	if (likely(bfqd->nonrot_with_queueing))
		return NULL;

	/*
	 * Prevent bfqq from being merged if it has been created too
	 * long ago. The idea is that true cooperating processes, and
	 * thus their associated bfq_queues, are supposed to be
	 * created shortly after each other. This is the case, e.g.,
	 * for KVM/QEMU and dump I/O threads. Basing on this
	 * assumption, the following filtering greatly reduces the
	 * probability that two non-cooperating processes, which just
	 * happen to do close I/O for some short time interval, have
	 * their queues merged by mistake.
	 */
	if (bfq_too_late_for_merging(bfqq))
		return NULL;

	if (bfqq->new_bfqq)
		return bfqq->new_bfqq;

	if (!io_struct || unlikely(bfqq == &bfqd->oom_bfqq))
		return NULL;

	/* If there is only one backlogged queue, don't search. */
	if (bfq_tot_busy_queues(bfqd) == 1)
		return NULL;

	in_service_bfqq = bfqd->in_service_queue;

	if (in_service_bfqq && in_service_bfqq != bfqq &&
	    likely(in_service_bfqq != &bfqd->oom_bfqq) &&
	    bfq_rq_close_to_sector(io_struct, request,
				   bfqd->in_serv_last_pos) &&
	    bfqq->entity.parent == in_service_bfqq->entity.parent &&
	    bfq_may_be_close_cooperator(bfqq, in_service_bfqq)) {
		new_bfqq = bfq_setup_merge(bfqq, in_service_bfqq);
		if (new_bfqq)
			return new_bfqq;
	}
	/*
	 * Check whether there is a cooperator among currently scheduled
	 * queues. The only thing we need is that the bio/request is not
	 * NULL, as we need it to establish whether a cooperator exists.
	 */
	new_bfqq = bfq_find_close_cooperator(bfqd, bfqq,
			bfq_io_struct_pos(io_struct, request));

	if (new_bfqq && likely(new_bfqq != &bfqd->oom_bfqq) &&
	    bfq_may_be_close_cooperator(bfqq, new_bfqq))
		return bfq_setup_merge(bfqq, new_bfqq);

	return NULL;