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SubjectRe: [PATCH v3] sctp: fix refcount bug in sctp_wfree
On Fri, Mar 20, 2020 at 07:09:59PM +0800, Qiujun Huang wrote:
> Do accounting for skb's real sk.
> In some case skb->sk != asoc->base.sk:
>
> for the trouble SKB, it was in outq->transmitted queue
>
> sctp_outq_sack
> sctp_check_transmitted
> SKB was moved to outq->sack

There is no outq->sack. You mean outq->sacked, I assume.

> then throw away the sack queue

Where? How?
If you mean:
/* Throw away stuff rotting on the sack queue. */
list_for_each_safe(lchunk, temp, &q->sacked) {
tchunk = list_entry(lchunk, struct sctp_chunk,
transmitted_list);
tsn = ntohl(tchunk->subh.data_hdr->tsn);
if (TSN_lte(tsn, ctsn)) {
list_del_init(&tchunk->transmitted_list);
if (asoc->peer.prsctp_capable &&
SCTP_PR_PRIO_ENABLED(chunk->sinfo.sinfo_flags))
asoc->sent_cnt_removable--;
sctp_chunk_free(tchunk);

Then sctp_chunk_free is supposed to free the datamsg as well for
chunks that were cumulative-sacked.
For those not cumulative-sacked, sctp_for_each_tx_datachunk() will
handle q->sacked queue as well:
list_for_each_entry(chunk, &q->sacked, transmitted_list)
cb(chunk);

So I don't see how skbs can be overlooked here.

> SKB was deleted from outq->sack
> (but the datamsg held SKB at sctp_datamsg_to_asoc

You mean sctp_datamsg_from_user ? If so, isn't it the other way
around? sctp_datamsg_assign() will hold the datamsg, not the skb.

> So, sctp_wfree was not called to destroy SKB)
>
> then migrate happened
>
> sctp_for_each_tx_datachunk(
> sctp_clear_owner_w);
> sctp_assoc_migrate();
> sctp_for_each_tx_datachunk(
> sctp_set_owner_w);
> SKB was not in the outq, and was not changed to newsk

The real fix is to fix the migration to the new socket, though the
situation on which it is happening is still not clear.

The 2nd sendto() call on the reproducer is sending 212992 bytes on a
single call. That's usually the whole sndbuf size, and will cause
fragmentation to happen. That means the datamsg will contain several
skbs. But still, the sacked chunks should be freed if needed while the
remaining ones will be left on the queues that they are.

Thanks,
Marcelo

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